If $A\subset \mathbb{N}$ has density $0$ then $s^{-1}(A)$ must also have density $0$.

Pollack, Pomerance, and Thompson [PPT18] prove that if $\epsilon(x)=o(1)$ and $A\subset \mathbb{N}$ has size at most $x^{1/2+\epsilon(x)}$ then $\#\{ n\leq x: s(n)\in A\} =o(x)$ as $x\to \infty$. It follows that (using $s(n)\ll n\log\log n$) if $A$ grows like $\lvert A\cap [1,x]\rvert\leq x^{1/2+o(1)}$ then $s^{-1}(A)$ has density $0$.