Let $\mathsf{d}^\star$ be the asymptotic upper density, defined on the power set of positive integers $\mathbf{N}^+$, so that
$$
\mathsf{d}^\star\colon \mathcal{P}(\mathbf{N}^+) \to\mathbf{R}\colon X\mapsto \limsup\_{n\to \infty} \frac{|X\cap [1,n]|}{n}.
$$
It is easy to verify that if $k\cdot \mathbf{N}^++h:=\{kx+h\colon x \in \mathbf{N}^+\}$ is an arithmetic progression of $\mathbf{N}^+$, and $X$ is a set of positive integers having no elements in common with $k\cdot \mathbf{N}^++h$, then
$$
\mathsf{d}^\star(X\cup (k\cdot \mathbf{N}^++h))=\mathsf{d}^\star(X)+\frac{1}{k}.
$$ 
The same reasoning can be extended to the upper analytic, upper logarithmic, upper Banach and upper Buck densities, at least. That's why one may ask if this property holds in general:


We say that a set function $\mu^\star\colon \mathcal{P}(\mathbf{N}^+)\to \mathbf{R}$ is an "*upper density*" whenever it satisfies the following axioms:


(F1) $\mu^\star(\mathbf{N}^+)=1$


(F2) $\mu^\star(X) \le \mu^\star(Y)$ if $X\subseteq Y$


(F3) $\mu^\star(X\cup Y) \le \mu^\star(X)+\mu^\star(Y)$


(F4) $\mu^\star(k\cdot X+h)=\mu^\star(X)/k$


for all positive integers $k,h$ and sets $X,Y \subseteq \mathbf{N}^+$. Then, is it true that if $X \cap (k\cdot \mathbf{N}^++h)=\emptyset$ then
$$
\mu^\star(X \cup (k\cdot \mathbf{N}^++h))=\mu^\star(X)+\frac{1}{k}\,\,?
$$



Additivity of upper densities with respect to arithmetic progressions of integers
The answer is in the negative. 


Let $f$ and $g$ be two upper densities (in the sense of the OP), and let $\alpha \in [0,1]$ and $q \in [1,\infty[$. Then the function 
$$h := (\alpha f^q + (1-\alpha) g^q)^{\frac{1}{q}}$$
is an upper density too (in particular, condition (F3) follows from Minkowski's inequality, which is why we need $q \ge 1$). 


Next, fix a set $X \subseteq 2\cdot\mathbf N^+$, let $x := 2f(X)$, $y := 2g(X)$ and $Y := 2 \cdot \mathbf N^+ + 1$, and suppose to a contradiction that $h$ is ``weakly additive'' (that is, $h(A \cup B) = h(A) + h(B)$ for all disjoint $A, B \subseteq \mathbf N^+$ such that $B$ is an (infinite) arithmetic progression), regardless of the actual values of the parameters $\alpha$ and $q$. Then, also $f$ and $g$ are weakly additive, and using that $f(Y) = g(Y)=\frac{1}{2}$, we obtain 
$$
\begin{split}
2h(X \cup Y) & = 2(\alpha (f(X \cup Y))^q + (1-\alpha) (g(X \cup Y))^q)^{\frac{1}{q}} \\ 
& = 2(\alpha (f(X) + f(Y))^q + (1-\alpha) (g(X) + g(Y))^q)^{\frac{1}{q}} \\
& = (\alpha (x+1)^q + (1-\alpha)(y+1)^q)^{\frac{1}{q}}
\end{split}
$$
and
$$
\begin{split}
h(X) + h(Y) & = (\alpha (f(X))^q + (1-\alpha) (g(X))^q)^{\frac{1}{q}} + \frac{1}{2} \\ & = (\alpha x^q + (1-\alpha)y^q)^{\frac{1}{q}}+ \frac{1}{2}
\end{split}
$$
which, together with $h(X \cup Y) = h(X) + h(Y)$, yields
$$ (\alpha (x+1)^q + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (\alpha x^q + (1-\alpha)y^q)^{\frac{1}{q}} + 1.
$$ 
On the other hand, an appropriate choice of $f$, $g$ and $X$ makes it possible to have $x$ equal to zero while $y$ takes any prescribed value in the interval $[0,1]$: This can be achieved, for instance, by letting $f$ be the upper asymptotic density (on $\mathbf N^+$), $g$ the upper Banach density, and $X$ a suitable subset of the intersection, $S$, of $\bigcup\_{n \ge 1} [\![2^n, 2^n + n]\!]$ and $2 \cdot\mathbf N^+$, and by considering that (i) the upper asymptotic density of $S$ is $0$, (ii) the upper Banach density of $S$ is $\frac{1}{2}$, (iii) the upper asymptotic and upper Banach densities are upper densities, and (iv) upper densities have the strong, and hence the weak, Darboux property (by the main theorem [here](http://arxiv.org/abs/1510.07473)).


Accordingly, we should have
$$(\alpha + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (1-\alpha)^{\frac{1}{q}}y + 1$$
for all $\alpha, y \in [0,1]$ and $q \in [1,\infty[$, which, however, is blatantly false. []


**Added later.** If you assume $\alpha = \frac{1}{2}$ and $q = 2$ in the last displayed equation, you don't even need to know that the upper Banach density has the weak Darboux property, since then you end up with the equation 
$$\sqrt{1 + (y+1)^2} = y + \sqrt{2},$$
which has a unique solution for $y \in \bf R$ (namely, $y = 0$).


