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Let $\Omega \subset R^2$ is a ball. Consider the equation $$ -\triangle u = f(x), \quad x \in \Omega $$ $$ u = 0 \quad x \in \partial\Omega. $$ It suffices to prove that for $p \geq 2$ $$ \|D^2u\|_{L^p(\Omega)} \leq C \|f\|_{L^p(\Omega)}. $$ At first, as you know, using integration by parts we have \begin{equation}\label{equ1} \|u\|_{H^1(\Omega)} \leq C \|f\|_{L^2} \leq C \|f\|_{L^p(\Omega)}. \end{equation} Then consider a cutoff function $\eta \in C^\infty_0(\Omega)$, denote by $v = \eta u$, then $v$ satisfies the equation $$ -\triangle v = \eta f - 2\nabla u \cdot \nabla \eta - \triangle \eta u, \quad x \in R^2. $$ It's known that $\xi_i\xi_j/|\xi|^2$ is an $L^{q}$ multiplier, that is, $$ \|\partial_i\partial_j u\|_{L^q(R^2)} \leq C \|\triangle u\|_{L^q(R^2)}, \quad q \in (1, \infty). $$ Using this fact and \eqref{equ1}, notice the support of $\eta$ we obtain $$ \|u\|_{H^2{(\Omega)}} \leq C \|f\|_{L^p}(\Omega). $$ Then the Sobolev embedding theorem yields that $$ \|u\|_{W^{1,q}{(\Omega)}} \leq C \|f\|_{L^p(\Omega)}, \quad 2 \leq q < \infty. $$ Proceed the above argument again, we find $$ \|D^2u\|_{L^p(\Omega)} \leq C \|f\|_{L^p(\Omega)} $$ as desired.