You can deduce this from the general theory, without explicitly constructing counterexamples. Assume that you had an estimate of the form $||u||_{H^2} \leq C(||u||_{L^2} + ||\Delta u||_{L^2})$ for $u \in H^2(\Omega)$. Consider the vector space $V = \{u \in H^2(\Omega) \, | \, \Delta u = 0 \}$ consisting of harmonic functions on $\Omega$ as a normed vector space with the $||\cdot||_{H^2}$ norm. Look at the closed unit ball $B$ of $(V, ||\cdot||_{H^2})$. The embedding $H^2(\Omega) \hookrightarrow L^2(\Omega)$ is compact so any sequence $u_n \in B$ has a subsequence $u_{n_k}$ that is Cauchy in $L^2(\Omega)$. The estimate $||u||_{H^2} \leq C ||u||_{L^2}$ shows that $u_{n_k}$ is also Cauchy in $H^2(\Omega)$ which implies that $B$ is compact. Thus $V$ must be a finite dimensional vector space. However, this is absurd as without imposing any boundary conditions there are infinitely many linearly independent harmonic functions in $H^2(\Omega)$ on any open subset where $n \geq 2$ (at least when $\Omega$ is bounded).
This argument generalizes in the sense that any time you have an estimate of the form $||x||_{X} \leq C(||Dx||_{Y} + ||Kx||_{Z})$ where $X,Y,Z$ are Banach spaces, $D \colon X \rightarrow Y$ is a bounded operator and $K \colon X \rightarrow Z$ is a compact operator it implies that $D$ is a semi-Fredholm operator, meaning it has a closed image and a finite dimensional kernel. In our example, $D = \Delta$ and $K$ is the compact inclusion $H^2(\Omega) \hookrightarrow L^2(\Omega)$. Since such estimates imply that the kernel of $D$ is finite dimensional, the spaces involved should already encode appropriate boundary value conditions for $D$. In our example, $X = H^2(\Omega) \cap H^1_0(\Omega)$ and the $H^1_0(\Omega)$ part handles the boundary value conditions.
