No. It is not true.
However, I was wondering under what conditions it is true.
I think that if $s>n/2$ and $1/f$ is bounded the result holds. Is it true?
Edit. If the domain is unbounded the set described may be empty as shown by TZakrevskiy's answer. Therefore, I am focusing on compact or bounded domains.
The obvious big question - can you find $f\in H^s$ such that $1/f$ is bounded?
Consider $\Bbb R$: if $1/f \in L^\infty(\Bbb R)$, then $|f|$ is bounded from below by $\frac{1}{vrai\sup |1/f|}>0$, therefore $f \notin L^s(\Bbb R)$ for $s\ge 1$.
Unless we answer these considerations rigorously, I suspect that your hypothesis describes an empty set of functions.