 Squarefree-part bound for smooth numbers

Let $S$ be a finite set of prime numbers and let $f_S(x)$ count S-smooth numbers $1\le n\le x$, where $x\ge 1$. Each such $n$ has a unique expression $n=a{b}^2$, where $a$ is a positive squarefree integer and $b$ is a positive integer. Split every prime exponent into its parity and an even remainder to obtain this decomposition.

Every prime factor of $a$ belongs to $S$, so $a$ is the product of a subset of $S$, giving at most $2^{|S|}$ possibilities. Also $b\le \sqrt{x/a}\le \sqrt{x}$. Ignoring further restrictions on $b$ only increases the count, provingIn particular, each fixed finite-prime family has natural density zero. This differs from allowing the smoothness bound, and hence the prime set, to grow with $x$.