Let $E := \{u \in H^1(\Omega) \mid \Delta u \in L^2(\Omega)\}$ be equipped with the norm: $$\|u\|_E := \sqrt{\|u\|_{L^2}^2 + \|\nabla u\|_{L^2}^2 + \|\Delta u\|_{L^2}^2}$$ Almost by definition, $E$ is a Banach space with said norm.
Per the answers to the question Do $u\in H^1(\Omega)$ and $\Delta u\in L^2(\Omega)$ imply $u\in H^2(\Omega)$?, $E$ is generally distinct from $H^2$ even if $\partial \Omega$ is smooth due to integrability-up-to-the-boundary issues. On the other hand, the inequality you desire would imply that $H^2$ is closed in $E$.
It thus suffices to check that $H^2$ is dense in $E$ to show that your inequality cannot exist when $E$ is distinct from $H^2$, since dense and closed would imply that $H^2 = E$.
Now the procedure is a bit involved due to having to deal with the boundary, but an adaptation of the proof of Theorem $3$ of paragraph $5.3.3.$ in Evans' Partial differential equations should yield that $C^\infty\left(\overline{\Omega}\right) \cap E$ is dense in $E$ whenever $\partial \Omega$ is $C^1$, yet $C^\infty\left(\overline{\Omega}\right) \cap E \subseteq H^2 \subset E$, hence $H^2$ is dense in $E$ in that case, and we are done.
A more self-contained and confident answer would write down all the modifications, so you can see this more as a comment-answer if you want, but essentially the reason this should work is that you can get and use estimates on one combination of partial derivatives at a time (in this instance, I'm referring to the Laplacian) without having to make use of all the derivatives at the same time or only being able to use triangle inequalities with estimates on the "elementary" partial derivatives (each second derivative here).
