Let $\Omega \subset R^2$ is a ball. Consider the equation $$ -\triangle u = f(x), \quad x \in \Omega $$ $$ u \big|_{\partial \Omega} = 0. $$
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
$$ \|u\|_{H^1(\Omega)} \leq C \|f\|_{L^2} \leq C \|f\|_{L^p(\Omega)}. $$
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 the above facts, 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.