Consider the Poisson's equation $$\nabla^2\phi(\textbf{x})=-\rho(\textbf{x})/\epsilon_0.$$ What is the meaning of the inverse operator in the following $$\phi(\textbf{x})=-\frac{1}{\nabla^2}\frac{\rho(\textbf{x})}{\epsilon_0}.$$
How do I show that $\frac{1}{\nabla^2}$ is equivalent to an integral operator acting on $\rho(\textbf{x})$?
Since you are using physical notation, it is reasonable to assume that your boundary condition is $\phi(\vec x) \to 0 $ as $|\vec x| \to \infty$. In this situation, the solution is $$ \phi(\vec x) = \frac 1 {4\pi \epsilon_0}\iiint d^3 \vec y \frac{\rho(\vec y)}{ |\vec x - \vec y|}.$$ So $$ - \frac 1 {\nabla^2} \frac{\rho(\vec x)}{\epsilon_0} = \frac 1 {4\pi \epsilon_0}\iiint d^3 \vec y \frac{\rho(\vec y)}{ |\vec x - \vec y|}.$$
As Ian mentioned in his comment, there is a rigorous proof of this in Evans, but, assuming you're a physicist, I would prefer to explain it like this:
The function $$ G(\vec x, \vec y) = \frac 1 {4\pi \epsilon_0 |\vec x - \vec y|}$$ is the electric potential around a point charge at position $\vec y$. In other words, it is the solution to $$ \nabla^2_{\vec x} G(\vec x, \vec y) = - \frac 1 {\epsilon_0} \delta^3(\vec x - \vec y), $$ obeying vanishing boundary conditions at infinity. [This is essentially Gauss' law from vector calculus. You can derive it by considering volume integrals of both sides of the PDE on spherical balls around $\vec y$.]
The solution I gave you can be thought of as a sum over the electric potential contributions from the infinitesimal charges $\rho(\vec y) d^3 \vec y$ at all possible positions $\vec y$.