integration - General Solution to Laplace's Equation in $mathbb{R^3}$

I am trying to find the Green's function for $\ \nabla^2\phi=S(x)\ $ for $\ x\in\mathbb{R^3}$ and express the general solution to Laplace's equation in $\mathbb{R^3}$.

To find the Green's function, I considered
\begin{align}

\nabla^2G&=\delta\left(\underline{x}-\underline{x'}\right) \\
\nabla^2G&=0 \ \ \ \ \ \ \left(\text{when} \ \underline{x}\neq\underline{x'}\right)\\
\nabla^2G&=\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{DG}{dr}\right) \\
G(r)&=-\frac{A}{r}+B, \ \ A,B\in\mathbb{R}.
\end{align}
Next consider a ball centred at $\underline{x'}$ with radius $a$, denoted $B_a$. Then,
\begin{align}
\iiint_{B_a} \nabla^2G \ dV&=\iiint_{B_a}\delta\left(\underline{x}-\underline{x'}\right) \ dV \\
\iint_{\delta B_a}\frac{dG}{dr}\bigg|_{r=a} ds &=1 \ \ \ \text{(Divergence theorem)}\\
4\pi A&=1 \\

\implies A&=\frac{1}{4\pi}
\end{align}
Hence (letting $B=0$ without loss of generality), $$G(r)=-\frac{1}{4\pi r}.$$ Now, $r=|\underline{x}-\underline{x'}|$, which means $$G(\underline{x},\underline{x'})=-\frac{1}{4\pi|\underline{x}-\underline{x'}|}.$$

Why can the general solution be expressed as $$\phi(\underline(x))=-\frac{1}{4\pi}\iiint_{\mathbb{R^3}}\frac{1}{|\underline{x}-\underline{x'}|} S(\underline{x'}) \ dV'?$$

Answer

The idea is to find $G$ such that $\nabla^2 G = \delta(x)$ in the distributional sense. Then let
$$\phi(x) := (G *S)(x)$$

Then because you have a convolution it follows that
$$\nabla^2 \phi = \nabla^2G*S = \delta * S = S$$
Where $\delta$ acts as an identity with respect to convolution and so $\phi$ satisfies Laplace's equation.

EDIT :

Instead of thinking of $G$ as a function of two variables $(x,x')\in \mathbb{R}^3\times\mathbb{R}^3$ you can write it as of function of $x$ only which gives
$$G(x) = \frac{-1}{4\pi|x|}$$
Which comes from setting $\nabla^2_x G(x) = \delta(x)$ instead of $\nabla^2_x G(x,x') = \delta(x -x')$. This means you integrate over balls centered at zero to construct $G$ instead. So that
$$G*S = \frac{-1}{4\pi}\int_{\mathbb{R}^3} \frac{1}{|x-x'|}S(x')dx'$$