Let $(Y,\left\|\cdot\right\|\_Y)$ be a Banach space and $A:D(A)\subset Y \to Y$ a closed operator. Studying dynamical systems of the form
\begin{equation}
u'=Au
\end{equation}
quickly leads to the notion of one-parameter semigroups:


$\textbf{Definition}$: A one-parameter semigroup $G$ is a collection $\left\{S(t):Y\to Y\right\}\_{t \geq 0}$ of bounded operators which satisfy $S(0)=I$ and $\forall s,t > 0$: $S(s+t)=S(s)S(t)$.
A one-parameter semigroup is called strongly continuous if the maps
\begin{equation}
t \mapsto S(t)u
\end{equation}
are continuous $\forall u \in Y$. 


My problem is the following: what if the nice $C^1$-solutions $u(t)$ ($u(t) \subset D(A)$) of the $(u'=Au)$-system have not for all $t>0$ a uniform bound 
\begin{equation}
\sup\_{u(0) \in D(A),\;u(0)\text{ generates $C^1$-solution}}\frac{\left\|u(t)\right\|\_Y}{\left\|u(0)\right\|\_Y}
\end{equation}
but nonetheless a vector subspace $X \subset Y$ exists with $D(A) \subset X$, together with some norm $\left\|\cdot\right\|\_X$ such that $(X,\left\|\cdot\right\|\_X)$ is a Banach space of its own and
\begin{equation}
\sup\_{u(0) \in D(A),\;u(0)\text{ generates $C^1$-solution}}\frac{\left\|u(t)\right\|\_Y}{\left\|u(0)\right\|\_X}
\end{equation}
does turn out to be finite for all $t>0$. Continuing this line of reasoning I end up with something like


$\textbf{Definition}$: Let $(Y,\left\|\right\|\_Y)$ and $(X,\left\|\right\|\_X)$ be Banach spaces and $X \subset Y$ such that $X$ is dense in $Y$ for the $Y$-norm. A bilateral one-parameter semigroup $G$ is a collection $\left\{S(t):X\to Y\right\}\_{t \geq 0}$ of bounded operators which satisfy:


\*$S(0)=I\_{X \to Y}$


\*$\forall u \in X$: $\varphi\_u:[0,\infty) \to Y:t \mapsto S(t)u$ is a continuous map.


\*$\forall u \in X$ and $f:[0,\infty) \to \mathbb{C}$ a $K\_G$-function (see below for definition), we have $\int\_0^{\infty} f(s)S(s)u\text{ d}s \in X$ and
\begin{equation}
S(t)\left(\int\_0^{\infty} f(s)S(s)u\text{ d}s\right) = \int\_0^{\infty} f(s)S(t+s)u\text{ d}t.
\end{equation}
$\textbf{Definition}$: The vector space $K\_G$ consists of all complex functions with domain $[0,\infty)$ spanned by functions of the form $\chi\_{[a,b]}$ or $f$ $C^1$ such that $\int\_0^{\infty} \left\|f'(s)S(s)\right\|\text{d}s < \infty$, $\forall t \geq 0$ $\int\_0^{\infty} \left\|f(t)S(s+t)\right\|\text{ d}s <\infty$, and $\left\|f(t)S(t)\right\| \to 0$ as $t\to \infty$. 
*I think we could as well replace "spanned by functions of the form $\chi\_{[a,b]}$ or $f$ $C^1$ such that..." in the definition of $K\_G$ by "functions which are piecewise $C^1$ up to a set of 0 Lebesgue-measure".*


Are there any good reasons to reject this line of reasoning and this generalisation from the start? Are my concerns (which seem generic to me) already addressed somewhere in the literature?


EDIT: an example and elaboration is below in an answer.



"Generalisation" of one-parameter semigroups
In fact, even a snake does not have to look beyond its nose to find a nice example (I post it here such that the original question does not get to long, discouraging potential readers).
Take $X$ as the complex-valued sequences with the usual vector-space structure and with norm
\begin{equation}
\left\|x\right\|\_X^2=|x\_1|^2+|x\_2|^2+|x\_3|^3...
\end{equation}
(i.e. $X=l^2(\mathbb{N})$)
and $Y$ again those sequences with the more relaxed norm
\begin{equation}
\left\|x\right\|\_Y^2=\frac{1}{2}(|x\_1|^2+\frac{1}{2!}|x\_2|^2+\frac{1}{3!}|x\_3|^2+...)
\end{equation}
For the role of closed operator $A:D(A)\subset Y \to Y$ we take the infinite diagonal matrix
\begin{equation}
A\_{kl}=\left(k+i\sqrt{2k!-k^2}\right)\delta\_{kl}.
\end{equation}
This seemingly "arbitrary" operator satisfies ($\forall x \in X$)
\begin{equation}
\left\|x\right\|\_X=\left\|Ax\right\|\_Y.
\end{equation}
which establishes that $D(A)=X$ (as a subset of $Y$ of course).


Next, let $Z\subset X$ be the sequences with finite support. $Z$ is dense in $X$ and $Y$ for their respective norms. It is straightforward to show that the sequence
\begin{equation}
u=(u\_1,...,u\_n,0,0...)
\end{equation}
is equal to $u(0)$ in the following solution of $u'=Au$ which is smooth in both the $X$ and $Y$-norm and unique for this initial condition:
\begin{equation}
u(t)=(u\_1\exp(A\_{11} t),...,u\_n \exp(A\_{kk}t),0,0...) (\in Z)
\end{equation}
It is not difficult to see that there is no uniform bound ($\forall u \in Y$ or $\in X$) to the quantities $\frac{\left\|u(t)\right\|\_Y}{\left\|u\right\|\_Y}$ or $\frac{\left\|u(t)\right\|\_X}{\left\|u\right\|\_X}$.


Assume however that $\left\|u\right\|\_X=1$ which means in particular $|u\_k|\leq 1$ $\forall k >0$. Then
\begin{equation}
\left\|u(t)\right\|\_Y = \frac{1}{2}\sum\_{k=1}^{\infty} \frac{1}{k!}|u\_k\exp(A\_{kk}t)|^2 \leq \frac{1}{2}\sum\_{k=1}^{\infty} \frac{1}{k!}\exp(kt)^2=\frac{1}{2}\exp(\exp(2t))
\end{equation}


**So far, I've produced all prerequisites stated in my question above, which I will show to be enough to produce a bilateral semigroup. That is, I've come up with a $u'=Au$-system which satisfies...**


**Definition** Let $(Y,\left\|\right\|\_Y)$ be a Banach space, a closed linear operator $A:D(A)\subset Y \to Y$ is called a "**Cauchy-operator**" if the set $Z\subset D(A)$, where $u\in Z$ i.f.f. a unique solution to $u'(t)=Au(t)$ ($\forall t \geq 0$) exists with initial condition $u$, is dense in $Y$. From now we define semigroup operators $R(t):Z \to D(A)$ through $u(t)=R(t)u$.


We call $A$ a "**regular Cauchy-operator**" if there exists moreover a vectorspace $X$ with $D(A) \subset X \subset Y$ with an accompanying norm $\left\|\right\|\_X$ such that $(X,\left\|\right\|\_X)$ is a Banach space and such that $Z$ is also dense in $(X,\left\|\right\|\_X)$ and such that $\sup\_{u \in Z}\frac{\left\|R(t)u\right\|\_Y}{\left\|u\right\|\_X} < \infty$ $\forall t >0$. $A$ is called a "**smooth Cauchy-operator**" if for every $t^\* \in [0,\infty)$, there exists $\epsilon>0$ such that 
\begin{equation}
\sup\_{u \in Z, t\in [t^\*-\epsilon,t^\*+\epsilon]}\frac{\left\|R(t)u\right\|\_Y}{\left\|u\right\|\_X} < \infty
\end{equation}


**Theorem** Let $A:D(A)\subset Y \to Y$ be a Cauchy operator. $Z$ is a vector space and the $R(t):Z \to D(A)$-operators are linear.
Their range is contained in $Z \subset D(A)$ and $\forall s \geq 0$. Redefining them as $R(t):Z\subset Y \to Z \subset Y$ we have
\begin{equation}
R(s+t)u=R(s)R(t)u.
\end{equation}
**proof**: exercise.


**Theorem** If $A:D(A)\subset Y \to Y$ is a **regular** Cauchy operator. Then the bounded operators $R(t):Z\subset X \to Z \subset Y$ can be continuously extended to bounded operators
\begin{equation}
S(t):X \to Y.
\end{equation}
If at some point $t^\*\in [0,\infty)$ and for some $\epsilon>0$, we have that $R(t)$ is uniformly bounded on $[t^\*-\epsilon,t^\*+\epsilon]$, then for all $u\in X$ the map
\begin{equation}
\varphi\_u:[0,\infty)\to Y: t \mapsto S(t)u
\end{equation}
is continuous at $t^\*$. As a consequence these maps are continuous if $A$ is a **smooth** Cauchy operator.
In this case the operators $S(t)$ constitute a bilateral one-parameter semigroup $G=\left\{S(t)\right\}\_{t\geq 0}$. **Note: I've edited the definition of bilateral semigroup slightly.**


**proof** The first part of the theorem is straightforward and left to the reader. Keep in mind that the continuity of the $\varphi\_u$ maps implies uniform continuity: for all compacts $C \subset [0,\infty)$ we have
\begin{equation}
\sup\_{t\in C} \left\|S(t)\right\| \leq 0.
\end{equation}
For the second part of the theorem, we first have to show that elements of the form $\int\_a^b S(s)u\text{d}s=\int\_0^{\infty} \chi\_{[a,b]}S(s)u\text{d}s$ or $\int\_0^{\infty} f(s)S(s)u\text{d}s$
(where $f:[0,\infty)\to \mathbb{C}$ is differentiable) are in $X$. Secondly, we have to show that "$S(t) \int S(s)... = \int S(s+t)...$". (Important note: all the relevant Bochner integrals are of course defined in the Banach space $Y$!) 
In this proof I'll only treat the second case, while the first is similar.


Take $(u\_n)\_n \in Z$ such that $\left\|u\_n-u\right\|\_X \to 0$. By the boundedness of the operators $S(t)$, we have
\begin{equation}
\left\|\int\_0^{\infty} f(s)S(s)u\_n\text{d}s - \int\_0^{\infty} f(s)S(s)u\text{d}s\right\|\_Y \to 0
\end{equation}
(exercise)
Moreover, we have 
\begin{equation}
\left\|A\int\_0^{\infty} f(s)S(s)u\_n\text{d}s + f(0)u+\int\_0^{\infty} f'(s)S(s)u\text{d}s\right\|\_Y = \left\|\int\_0^{\infty} f(s)AS(s)u\_n\text{d}s + f(0)u+\int\_0^{\infty} f'(s)S(s)u\text{d}s\right\|\_Y =\left\|\int\_0^{\infty} f(s)(S(s)u\_n)'\text{d}s + f(0)u+\int\_0^{\infty} f'(s)S(s)u\text{d}s\right\|\_Y=\left\|-f(0)u\_n - \int\_0^{\infty} f'(s)S(s)u\_n\text{d}s+ f(0)u+\int\_0^{\infty} f'(s)S(s)u\text{d}s\right\|\_Y \to 0
\end{equation}
So the closedness of $A:D(A) \subset Y \to Y$ implies that $v:=\int\_0^{\infty} f(s)S(s)u\text{d}s \in D(A)\subset X$ and $Av =-f(0)u-\int\_0^{\infty} f'(s)S(s)u\text{d}s$. To finish, one takes the following steps:
\*$\eta:[0,\infty)\to Y: t \mapsto \int\_0^{\infty} f(s)S(s+t)u\_n\text{d}s=\int\_t^{\infty} f(s-t)S(s)u\_n\text{d}s$ is differentiable and $\eta'(t)=A\eta(t)$. Hence $\eta(t) \in Z$ for all $t\geq 0$.


\*But then
\begin{equation}
\nu:[0,\infty)\to Y: h \mapsto S(t-h)\int\_0^{\infty} f(s)S(s+h)u\_n\text{d}s = S(t-h)\eta(h)
\end{equation}
is also differentiable and $\nu'(h)=0$. But then
\begin{equation}
S(t)\int\_0^{\infty} f(s)S(s)u\_n\text{d}s = \nu(0)=\nu(t)=\int\_0^{\infty} f(s)S(s+t)u\_n\text{d}s.
\end{equation}


