Let $f : \mathbb{R} \to \mathbb{R}$ be a Lebesgue measurable function and $D$ be a countable dense subset of $\mathbb{R}$. Suppose that for a.e. $x \in \mathbb{R}$ we have \begin{equation*} f(x + d) = f(x) \qquad \text{for every } d \in D \end{equation*} (notice that, without loss of generality, we can assume that $D$ is a module over the integers).
Does it follow that $f(x) = c$ a.e.?
The answer is yes.
Indeed, in view of the possible replacement of $f(x)$ by $\min(N,\max(-N,f(x))$ for an arbitrary real $N>0$ and all real $x$, assume without loss of generality that the function $f$ is bounded.
Let then $$F(x):=\int_0^x f(t)\,dt$$ for real $x\ge0$ and $$F(x):=-\int_0^{-x} f(-t)\,dt$$ for real $x<0$. Then for all $d\in D$ and real $x\ge0$ $$F(x)=\int_0^x f(t+d)\,dt=F(x+d)-F(d),$$ so that $$F(x+d)=F(x)+F(d). \tag{1}\label{1}$$ Similarly, \eqref{1} holds for all real $x<0$ and all $d\in D$. Therefore and because $F$ is continuous, \eqref{1} holds for real $x$ and all real $d$. So, $F(x)=cx$ for some real $c$ and all real $x$. So, $f$, being an a.e. derivative of the absolutely continuous function $F$, equals $c$ a.e. $\quad\Box$
Yes. Suppose that $f(x)$ is not a.e. constant. Then there is some subset $X$ of $\mathbb{R}$ such that $A:=f^{-1}(X)$ has positive measure but not full measure (we can take $X=(-\infty,x]$ for some adequate $x\in\mathbb{R}$).
Note that for all $d\in D$, $\mu(A\mathbin\Delta(A-d))=0$. Also, as $A$ and $\mathbb{R}\setminus A$ have positive measure, by the Lebesgue density theorem there are two points $x,y$ such that, for some small value of $\varepsilon$, $\mu(A\cap(x-\varepsilon,x+\varepsilon))>1.5\varepsilon$ and $\mu(A\cap(y-\varepsilon,y+\varepsilon))<0.5\varepsilon$.
However, there is a sequence $(d_n)_n$ in $D$ such that $x+d_n\to y$. And for all $n$ we have $$\mu(A\cap(x+d_n-\varepsilon,x+d_n+\varepsilon))=\mu((A-d_n)\cap(x-\varepsilon,x+\varepsilon))$$ $$=\mu(A\cap(x-\varepsilon,x+\varepsilon))>1.5\varepsilon.$$
Thus, $\mu(A\cap(y-\varepsilon,y+\varepsilon))=\lim_{n\to\infty}\mu(A\cap(x+d_n-\varepsilon,x+d_n+\varepsilon))\geq1.5\varepsilon$, a contradiction.
Edit: It seems this argument works more generally, e.g. if $G$ is a locally compact group with $\sigma$-finite Haar measure and a measurable function $f:G\to\mathbb{R}$ satisfies that for all $d$ in some subset $D$ dense in $G$ we have $f(dx)=f(x)$ a.e., then $f$ is a.e. constant. To use the argument above in the general case one would need to use some version of the Lebesgue density theorem for locally compact groups, see e.g. Theorem A in the article "Three Results for Locally Compact Groups Connected with the Haar Measure Density", by Mueller.
Start as in Iosif's answer and observe that it suffices to discuss the case when $f$ is bounded. Then $f$ is locally integrable, so defines a distribution $f\in\mathcal D'$. The distributional derivative may be computed as $f'=\lim_{h\to 0} (f(x+h)-f(x))/h$, with the limit taken in $\mathcal D'$ (that is, we have convergence in $\mathbb C$ after applying both sides to an arbitrary test function). By taking it along a sequence $h\in D$, we see that $f'=0$, hence $f=c$.
