It is sufficient to require $u\in W^{1,\infty}(\Omega)\cap H^1_0(\Omega)$. Suppose this. Take $v\in H^{-1}(\Omega)$ and $w\in H^1(\Omega)$.
The notation $uv$ is kind of sloppy, I guess you meant something like
$$
(uv)(w) := v(uw) \quad w\in H^1_0(\Omega).
$$
Then
$$
|uv(w)| = |v(uw)| \le \|v\|_{H^{-1}} \|uw\|_{H^1(\Omega)} \le \|v\|_{H^{-1}}
( \|u\|_{L^\infty(\Omega)}\|\nabla w\|_{L^2(\Omega)} + \|\nabla u\|_{L^\infty(\Omega)}\| w\|_{L^2(\Omega)} )
\le 2\|v\|_{H^{-1}} \|u\|_{W^{1,\infty}} \|w\|_{H^1}.
$$
This proves, that the mapping $w\mapsto v(uw)$ is bounded.