Can multiples of two reals stay separated?

For any positive irrational $r$, the Beatty sequence is the sequence of numbers

$$\mathcal{B}_r = \lfloor r \rfloor, \lfloor 2r \rfloor, \lfloor 3r \rfloor, \ldots$$

Rayleigh theorem (also known as Beatty's theorem) states that

Given any irrational $r > 1$, there exists irrational $s > 1$ so that $\mathcal{B}_r$ and $\mathcal{B}_s$ partition the set of positive integers. i.e. each positive integer belongs to exactly one of the two sequences.

For this to happen, the condition is $$\frac1r + \frac1s = 1 \iff s = \frac{r}{r-1}$$ Translate this to the problem at hand. For irrational $x > 1$, just take $y = \frac{x}{x-1}$ and we will have

$$S(x) \cup S(y) = \mathbb{N}\quad\text{ and }\quad S(x) \cap S(y) = \{ 0 \}$$

For more details, see the wiki entry for Beatty's sequences and refs there.