Let $H$ be a Hamel basis; that is, $H$ is a set of real numbers such that every real number $x$ can uniquely be written in the form
$$x = \sum_{h \in H} q(x, h) \cdot h,$$
where the $q(x, h)$ are all rational and vanish for all but a finite number (depending on $x$) of $h$'s. The existence of Hamel bases can be proved via Zorn's lemma or Zermelo's well ordering theorem or any other statement equivalent to the axiom of choice.

We are now going to prove that the set $S$ of those real numbers $x$ whose $q(x, h)$ in $(*)$ are all integral satisfies the required condition.

To this end, fix a real number $x$. Since the conclusion is clear if $x = 0$, let $x$ be different from $0$ and let $m(x)$ be the least common multiple of the denominators of the non-vanishing $q(x, h)$ in $(*)$. Finally, notice that $m(x) \cdot x$ is a member of $S$, to conclude that any set of the form $nx + S$, where $n$ is a non-negative integer, must be one of the sets $rx + S$, $r = 0, 1, \dots, m(x) - 1$.