Let


* $(E,\mathcal E)$ be a measurable space
* $\mathcal E\_b:=\left\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\right\}$
* $(\kappa\_t)\_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$
* $Q$ denote the weak generator of $(\kappa\_t)\_{t\ge0}$; i.e. $$\mathcal D(Q):=\left\{f\in\mathcal E\_b\mid\forall x\in E:[0,\infty)\ni t\mapsto(\kappa\_tf)(x)\text{ is right-differentiable at }0\right\}$$ and $$(Qf)(x):=\left.\frac{\rm d}{{\rm d}t}(\kappa\_tf)(x)\right|\_{t=0+}\;\;\;\text{for }x\in E\text{ and }f\in\mathcal D(Q)$$
* $(\Omega,\mathcal A,\operatorname P)$ be a probability space
* $(Y\_t)\_{t\ge0}$ be an $(E,\mathcal E)$-valued time-homogeneous Markov process on $(\Omega,\mathcal A,\operatorname P)$ with transition semigroup $(\kappa\_t)\_{t\ge0}$
* $\alpha$ be a transition kernel on $(E,\mathcal E)$ and $$Af(x):=\int\_E ( f(y)-f(x)) \:\alpha(x,{\rm d}y)\;\;\;\text{for all }x\in E\text{ and }f\in\mathcal D(A):=\mathcal E\_b$$



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> **Question**: How can we construct an $(E,\mathcal E)$-valued time-homogeneous Markov process $(X\_t)\_{t\ge0}$ on $(\Omega,\mathcal A,\operatorname P)$ with weak generator $$Lf=Qf+Af\;\;\;\text{for all }f\in\mathcal D(L)\subseteq\mathcal D(Q)\cap\mathcal D(A)?\tag1$$
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The idea is that the local behavior between jumps of $(X\_t)\_{t\ge0}$ is described by $(Y\_t)\_{t\ge0}$ and, assuming that $\alpha(x,B)=c(x)\mu(x,B)$ for all $x\in E$ for some $\mathcal E$-measurable $c:E\to[0,\infty)$ and a Markov kernel $\mu$ on $(E,\mathcal E)$, the jumps occur at a state-dependent rate $c$ and are performed according to the state-depedendent distribution $\mu$.


The process should be described by something like $$X\_t=\sum\_{n\in\mathbb N\_0}1\_{[\tau\_n,\:\tau\_{n+1})}(t)Y^{(n)}\_{t-\tau\_n}\;\;\;\text{for all }t\ge0\tag1,$$ where $\tau\_n$ is the time of the $n$th-jump and the $Y^{(n)}$ are independent copies of $Y$.



> 
> However, how do we need to define the $\tau\_n$ precisely and how do we see that the weak generator of $(1)$ is actually equal to $L$?
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I'm aware of the following simpler result: If $(W\_n)\_{n\in\mathbb N\_0}$ is a time-homogeneous Markov chain on $(\Omega,\mathcal A,\operatorname P)$ with transition kernel $\kappa$ and $(N\_t)\_{t\ge0}$ is a Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $r>0$ and $W$ is independent of $N$, then $$Z\_t:=W\_{N\_t}\;\;\;\text{for }t\ge0$$ is a time-homogeneous Markov process with transition semigroup $\left(e^{t(\kappa-r)}\right)\_{t\ge0}$ and generator $r\left(\kappa-\operatorname{id}\_{\mathcal E\_b}\right)$.


In particular, if $W$ is a random walk with step distribution $\alpha^{-1}\nu$; i.e. $W\_n=\sum\_{i=1}^n\xi\_i$ for all $n\in\mathbb N$ for some independent identically $\alpha^{-1}\nu$-distributed process $(Z\_n)\_{n\in\mathbb N}$ on $(\Omega,\mathcal A,\operatorname P)$, then the generator of $Z$ is given by $$\mathcal E\_b\ni g\mapsto\int g(\;\cdot\;+y)-g\:\nu({\rm d}y).$$


Maybe a similar construction and hence an expression different from $(1)$ from which it is easier to derive the desired result is possible in the setting of this question.



Construction of a Markov process with prescribed local behavior and state-dependent jump distribution
The construction given by the OP is almost correct. Here is a slight correction: $$
X\_t = \sum\_{n=0}^{\infty} 1\_{[\tau\_n,\tau\_{n+1})}(t) Y\_{t - \tau\_n}^{n} \;, \tag{1}
$$ where we have introduced


* $\{\tau\_i\}$ are a sequence of jump times defined via $\tau\_{i+1}=\tau\_i+\xi\_i$, $\tau\_0=0$, and $\{\xi\_i \} \overset{i.i.d.}{\sim} \operatorname{Exp}(1)$ ; and,
* $\{Y^{i}\}$ are independent realizations of $Y$ with $Y\_0^i=x$ if $i=0$ and else sample $Y\_0^i \mid (Y^0, \dots, Y^{i-1}, \xi\_0, \dots, \xi\_i) \sim \alpha(Y\_{\tau\_i - \tau\_{i-1}}^{i-1}, \cdot)$ .


In other words,
$$
X\_t = \begin{cases}
Y^0\_t & t < \tau\_1 \;, \\
Y^1\_{t-\tau\_1} & \tau\_1 \le t < \tau\_2 \;, \\
Y^2\_{t-\tau\_2} & \tau\_2 \le t < \tau\_3 \;, \\
\vdots
\end{cases}
$$


To see that the weak generator of (1) is indeed $L=Q+A$, write $f (X\_t) - f(x) = \rm{I} + \rm{II} + \rm{III}$ where \begin{align\*}
\rm{I} &:= (f(X\_t) - f(x)) 1\_{\{t < \tau\_1 \}} \;, \\
\rm{II} &:= (f(Y\_0^1) - f(Y\_{\tau\_1}^0)) 1\_{\{t \ge \tau\_1 \}} \;, \\
\rm{III} &:= (f(Y\_{\tau\_1}^0) - f(x) + f(X\_t) - f(Y\_0^1)) 1\_{\{t \ge \tau\_1 \}} \;.
\end{align\*}
Then \begin{align\*}
E[\rm{I}] &= e^{-t} ( \kappa\_t f(x) - f(x) ) = e^{-t} E \int\_0^t Qf (Y\_s^0) ds \;, \\
E[{\rm II} \mid \tau\_1 = s] &= E[f(Y\_0^1) - f(Y\_{s}^0)] 1\_{\{ t \ge s \}} = E[ A f(Y\_s^0) ] 1\_{\{ t \ge s \}} \;, \\
E[ \rm{II} ] &= E \int\_0^{\infty} E[ {\rm II} \mid \tau\_1 = s] e^{-s} ds = E \int\_0^t e^{-s} A f(Y\_s^0) ds \;.
\end{align\*} One can similarly show that $E( \rm{III} )$ is $O(t^2)$ for $t \in [0,1]$. Therefore, combining the above and using $(e^{-s} - e^{-t}) \le (t-s)$ for $t \ge s$, one obtains that for all $ t \in [0,1]$ $$
E[f(X\_t)] - f(x) = E\int\_0^t (A f(Y\_s^0) + Q f(Y\_s^0)) ds + O(t^2) \;.
$$
While this construction/analysis covers the case of constant jump rates, the case of state-dependent jump rates can be treated similarly as discussed in the comments below.


