I have read that we can define Laplace operator $-\Delta: D(-\Delta) \subseteq H_0^1(\Omega) \to L^2(\Omega)$, with $\Omega \subseteq \mathbb{R}^n$ a bounded open set and $$D(-\Delta):=\{ u \in H_0^1 : \exists f \in L^2 (\Omega) \mbox{ such that } u \mbox{ is a weak solution of } (1) \},$$ where
$$ -\Delta u=f \mbox{ in } \Omega, \ \ u=0 \mbox{ on } \partial\Omega \ \ \ \ \ \ (1). $$
If $\Omega$ is an open bounded set of class $C^1$, can I say that $D(-\Delta)=H_0^1(\Omega)$? How can I show that $ -\Delta$ is invertible?
You cannot say that if $u \in H^1_0$ then $\exists f \in L^2$ such that $-\Delta u=f$. You know that $-\Delta u \in H^{-1}$ but nothing more.
Take for example $\Omega=(-1,1)$ and $u(x)=|x|-1$. Then $u \in H^1_0$ but $-\Delta u \not \in L^2$.
To show that $-\Delta$ is invertible you need to show that for any $f \in L^2$ there exist $u$ satisfying $(1)$.
Since $(1)$ can be written as: $$\int_\Omega \nabla u \cdot \nabla v =\int_\Omega f v \ \ \forall v \in H^1_0$$
this is an application of functional analysis theorems (for example Poincaré inequality and Lax-Milgram).