Suppose that $\rho: \mathbb R^3 \to \mathbb R$ is a function that tells us the electric charge density at each point in space. According to Coulomb's law, the electric field at a point $x \in \mathbb R^3$ is $$ \tag{1} E(x) = \frac{1}{4 \pi \epsilon_0} \int_{\mathbb R^3} \rho(y) \frac{x - y}{\|x - y \|^3} \, dy. $$ Question: What is the easiest way to show rigorously that the function $E$ defined above satisfies the integral form of Gauss's law: if $\Omega \subset \mathbb R^3$ is a smooth $3$-manifold with boundary, then $$ \tag{2} \int_{\partial \Omega} E \cdot dS = \frac{1}{\epsilon_0} \int_\Omega \rho(x) \, dx. $$ Is it necessary to first show that $$ \tag{3} \nabla \cdot E = \frac{\rho}{\epsilon_0}, $$ or is there a way to show that (1) implies (2) directly without going through (3)?
Comments: Physics textbooks typically argue that (2) holds for the electric field due to a point charge, then argue that (2) follows for the field due to a general charge distribution by superposition. Once (2) is established, they derive (3) as a corollary using the divergence theorem. So the chain of logic is $$ (1) \implies (2) \implies (3). $$ This is a highly intuitive but non-rigorous argument. I wonder if there's a way to make it rigorous.
A rigorous proof that (3) follows from (1) can be found here. And it is easy to get from (3) to (2) using the divergence theorem. But I wonder if that is the most direct route from (1) to (2). Notably, that chain of reasoning $$ (1) \implies (3) \implies (2). $$ differs from the highly intuitive physics textbook approach.
Since this question is about a rigorous proof of a mathematical fact, I believe it's a good fit for math.stackexchange.
