Thanks to a conversation I have had with user TrialAndError, whom I deeply thank, here, I think I have been able to understand a proof for the particular case where $\boldsymbol{F}:\mathring{A}\subset\mathbb{R}^3\to\mathbb{R}^3$ is such that $\exists \boldsymbol{G}\in C^3(\mathring{A}):\boldsymbol{F}=\nabla^2 \boldsymbol{G}$ and $V$, with $ \bar{V}\subset\mathring{A}$, satisfies the assumptions of the divergence theorem, with $\boldsymbol{x}\in\mathring{V}$.
Then, this result shows, with an application of Leibniz's rule for differentiation under the integral sign, that
$$\boldsymbol{F}(\boldsymbol{x})=-\frac{1}{4\pi}\nabla^2\int_V\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$where $\mu'$ is the usual $3$-dimensional Lebesgue measure. The same result shows that the integral in the expression above belongs to $C^2(\mathbb{R}^3)$ and therefore a known identity for the curl of the curl means that$$\boldsymbol{F}(\boldsymbol{x})=\frac{1}{4\pi}\nabla\times\left[\nabla\times\int_V\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'\right]-\frac{1}{4\pi}\nabla\left[\nabla\cdot\int_V\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'\right]$$where the following
Lemma. Let $\varphi:V\subset\mathbb{R}^3\to\mathbb{R}$ be bounded and $\mu'$-measurable, with $\mu'$ as the usual $3$-dimensional Lebesgue measure, where $V$ is bounded and measurable (according to the same measure). Let us define, for all $\boldsymbol{x}\in\mathbb{R}^3$, $$\Phi(\boldsymbol{x}):=\int_V \frac{\varphi(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$then $\Phi\in C^1(\mathbb{R}^3)$ and, for $k=1,2,3$, $$\forall\boldsymbol{x}\in\mathbb{R}^3\quad\quad\frac{\partial \Phi(\boldsymbol{x})}{\partial x_k}=\int_V\frac{\partial}{\partial x_k} \left[\frac{\varphi(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'=\int_V \varphi(\boldsymbol{x}')\frac{x_k'-x_k}{\|\boldsymbol{x}-\boldsymbol{x}'\|^3}d\mu'$$
which is proved here allows the commutation between derivative and integral sings to get$$\boldsymbol{F}(\boldsymbol{x})=\frac{1}{4\pi}\nabla\times\int_V\nabla\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'-\frac{1}{4\pi}\nabla\int_V\nabla\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'$$$$=\frac{1}{4\pi}\nabla\times\int_V\nabla\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\times\boldsymbol{F}(\boldsymbol{x}')d\mu'-\frac{1}{4\pi}\nabla\int_V\nabla\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\cdot\boldsymbol{F}(\boldsymbol{x}')d\mu'$$$$=-\frac{1}{4\pi}\nabla\times\int_V\nabla'\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\times\boldsymbol{F}(\boldsymbol{x}')d\mu'+\frac{1}{4\pi}\nabla\int_V\nabla'\left[\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]\cdot\boldsymbol{F}(\boldsymbol{x}')d\mu'$$$$=-\frac{1}{4\pi}\nabla\times\int_V\nabla'\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]-\frac{\nabla'\times \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$$$+\frac{1}{4\pi}\nabla\int_V\nabla'\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]-\frac{\nabla'\cdot \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$$$=-\frac{1}{4\pi}\nabla\times\int_V\nabla'\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'+\frac{1}{4\pi}\nabla\times\int_V\frac{\nabla'\times \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$$$+\frac{1}{4\pi}\nabla\int_V\nabla'\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'-\frac{1}{4\pi}\nabla\int_V\frac{\nabla'\cdot \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$where an application of the divergence theorem gives
$$\int_V\nabla'\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'=\lim_{\delta\to 0}\int_{V\setminus B(\boldsymbol{x},\delta)}\nabla'\times\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]dx_1'dx_2'dx_3'$$$$=\lim_{\delta\to 0}\int_{\partial(V\setminus B(\boldsymbol{x},\delta))}\frac{\hat{\boldsymbol{n}}(\boldsymbol{x}')\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'=\int_{\partial V}\frac{\hat{\boldsymbol{n}}(\boldsymbol{x}')\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'$$
because the integral on the internal surface is $\int_{\partial B(\boldsymbol{x},\delta)}\frac{-\hat{\boldsymbol{n}}(\boldsymbol{x}')\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'\xrightarrow{\delta\to 0} \mathbf{0}$ and analogously$$\int_V\nabla'\cdot\left[\frac{\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}\right]d\mu'=\int_{\partial V}\frac{\boldsymbol{F}(\boldsymbol{x}')\cdot \hat{\boldsymbol{n}}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'$$
and therefore $$\boldsymbol{F}(\boldsymbol{x})=-\frac{1}{4\pi}\nabla\times\int_{\partial V}\frac{\hat{\boldsymbol{n}}(\boldsymbol{x}')\times\boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'+\frac{1}{4\pi}\nabla\times\int_V\frac{\nabla'\times \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$$$+\frac{1}{4\pi}\nabla\int_{\partial V}\frac{\boldsymbol{F}(\boldsymbol{x}')\cdot \hat{\boldsymbol{n}}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}dS'-\frac{1}{4\pi}\nabla\int_V\frac{\nabla'\cdot \boldsymbol{F}(\boldsymbol{x}')}{\|\boldsymbol{x}-\boldsymbol{x}'\|}d\mu'$$as we wished to prove.