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.