The existence and uniqueness of the pure Neumann boundary value problem for smooth data can be proved using the Green representation formula, explicitly.
First notice that for $-\Delta \Phi = \delta(x)$, by Green's second identity and convolution formula: for $x\in \Omega$
$$
\int_{\Omega} \Big(u(y) \Delta \Phi(y-x) - \Phi(y-x)\Delta u(y)\Big) dy = \int_{\partial \Omega}\left(u(y)\frac{\partial \Phi}{\partial n}(y-x) - \Phi(y-x)\frac{\partial u}{\partial n}(y) \right) dS(y),
$$
we have
$$
u(x) = -\int_{\Omega} \Phi(y-x)\Delta u(y) \, dy + \int_{\partial \Omega}\left(\color{red}{\Phi(y-x)}\frac{\partial u}{\partial n}(y) - u(y)\color{blue}{\frac{\partial \Phi}{\partial n}(y-x)} \right) dS(y).
$$
For Dirichlet problem, a correction function is constructed so that red term vanishes.
For Neumann problems, Green function is constructed so that $u(x)$ on each point is unique up to a constant. Some people prefer to set:
$$
\left\{\begin{aligned}
-\Delta \Phi &= \delta(x) \quad \text{in }\Omega,
\\
\frac{\partial \Phi}{\partial n} &= \frac{1}{|\partial \Omega|} \quad \text{on }
\partial\Omega,
\end{aligned}\right.
$$
where absolute value just means the Lebesgue measure of codimension $1$, i.e., surface area, so that the constant difference is the average of $u$ on $\partial \Omega$. I myself prefer to set something like:
$$
\left\{\begin{aligned}
-\Delta \Phi &= \delta(x) - \frac{1}{|\Omega|}\quad \text{in }\Omega,
\\
\frac{\partial \Phi}{\partial n} &=0 \quad \text{on }
\partial\Omega,
\end{aligned}\right.\tag{$\star$}
$$
so that you have:
$$
u(x) = \frac{1}{|\Omega|}\int_{\Omega}u(y)dy -\int_{\Omega} \Phi(y-x)\Delta u(y) \, dy + \int_{\partial \Omega}
\Phi(y-x)\frac{\partial u}{\partial n}(y)dS(y).
$$
This means the solution is also unique up to a constant, that constant is the average of $u$ in $\Omega$, in other words, once we fix this average, $u$ is pinned down to just one function. The existence of the Green function is another story, which relies on functional analysis, e.g. see here.
The unique of $u$ up to a constant can be either proved by maximum principle or energy method, but for Helmholtz decomposition does not address the uniqueness, we don't have to address this issue here.
Now for your equation, we know that the compatibility condition is satisfied because:
$$
\int_{\Omega}\Delta u = \int_{\Omega}\nabla \cdot u = \int_{\partial \Omega} \frac{\partial u}{\partial n} dS = \int_{\partial \Omega} E\cdot n\,dS.
$$
Consider $u$ is smooth and $\int_{\Omega} u = 0$, then
$$
u(x) = -\int_{\Omega} \Phi(y-x) (\nabla_y \cdot E(y)) \, dy + \int_{\partial \Omega}
\Phi(y-x) (E(y)\cdot n) dS(y).
$$
Here are another two proofs for smooth vector field. The Neumann boundary approach is good for less regular vector field (say the one has component in some Sobolev space), for smooth vector field, using vector potentials would be more preferable (at least for me).
The first one is using a pointwisely hold formula if $\Omega$ is convex with respect to a point $x_0 \in \mathbb{R}^3$:
$$
E = \nabla \phi + A,
$$
where
$$
A = -(x - x_0)\times \int^1_0 t \nabla \times E\big(x_0 + t(x- x_0)\big)\,dt ,
$$
and
$$
\phi = (x - x_0)\cdot \int^1_0 tE\big(x_0 + t(x- x_0)\big).
$$
Though this does not bear the exact same form, but the formula gives you more physical intuition. For reference please see here.
The second one is using the Dirichlet problem's green function. Now we want to solve a vector Poisson equation:
$$
\left\{\begin{aligned}
-\Delta F &= E\quad \text{in }\Omega,
\\
F &=0 \quad \text{on }
\partial\Omega,
\end{aligned}\right.
$$
Then for the Dirichlet Green function:
$$
\left\{\begin{aligned}
-\Delta \Phi &= \delta(x) \quad \text{in }\Omega,
\\
\Phi &=0 \quad \text{on } \partial\Omega.
\end{aligned}\right.
$$
Use the Green representation formula component-wise, we have
$$
F_i(x) = -\int_{\Omega} \Phi(y-x)\Delta F_i(y) \, dy - \int_{\partial \Omega} F_i(y) \frac{\partial \Phi}{\partial n}(y-x) dS(y),
$$
and this is
$$
F_i(x) = \int_{\Omega} \Phi(y-x)E_i(y) \, dy.
$$
Now use the vector Laplacian's identity for smooth vector fields
$$
E(x) = -\Delta F(x) = \nabla \times \nabla \times F - \nabla \nabla \cdot F
\\
=\nabla_x \times \left(\nabla_x \times \int_{\Omega} \Phi(y-x)E (y) \, dy\right) - \nabla _x\left(\nabla_x \cdot \int_{\Omega} \Phi(y-x)E(y) \, dy\right),
$$
where the subscript $x$ means taking gradient/div/curl w.r.t. $x$. Further moving the derivative into the integral:
$$
E(x) = \nabla_x \times \left(\int_{\Omega} \nabla_x \times \Big(\Phi(y-x)E (y)\Big) \, dy\right) - \nabla_x\left(\int_{\Omega} \nabla_x \cdot \Big(\Phi(y-x)E(y)\Big) \, dy\right)
\\
= \nabla_x \times \left(\int_{\Omega} \nabla_x \Phi(y-x)\times E (y) \, dy\right) - \nabla_x\left(\int_{\Omega} \nabla_x \Phi(y-x)\cdot E(y) \, dy\right)
\\
= - \nabla_x \times \left(\int_{\Omega} \nabla_y \Phi(y-x)\times E (y) \, dy\right) + \nabla_x\left(\int_{\Omega} \nabla_y \Phi(y-x)\cdot E(y) \, dy\right)
\\
= \nabla_x \times \left(\int_{\Omega} \Big( \Phi(y-x) \nabla_y \times E (y) - \nabla_y \times (\Phi(y-x)E(y))\Big) \, dy\right)
\\
+ \nabla_x\left(\int_{\Omega} \Big( -\Phi(y-x) \nabla_y \cdot E (y) + \nabla_y \cdot (\Phi(y-x)E(y))\Big) \, dy\right)
\\
= \nabla_x \times \left(\int_{\Omega} \Phi(y-x) \nabla_y \times E (y)\,dy - \int_{\partial \Omega} \Phi(y-x)n\times E(y)\,dS(y) \right)
\\
+ \nabla_x\left(-\int_{\Omega} \Phi(y-x) \nabla_y \cdot E (y) \, dy +\int_{\partial \Omega}\Phi(y-x)E(y)\cdot n\,dS(y) \right).
$$
So that: in $\Omega$
$$
E = \nabla \phi + \nabla \times A,
$$
where
$$
\phi(x) = -\int_{\Omega} \Phi(y-x) \nabla_y \cdot E (y) \, dy +\int_{\partial \Omega}\Phi(y-x)E(y)\cdot n\,dS(y),
\\
A(x) = \int_{\Omega} \Phi(y-x) \nabla_y \times E (y)\,dy - \int_{\partial \Omega} \Phi(y-x)n\times E(y)\,dS(y).
$$
You do not need to find such a set $K \subset \Omega$!
At first, let us assume that $\Omega \subset \mathbb{R}^d$ is bounded. Then there is an open ball $B_R(0) \subset \mathbb{R}^d$ such that $\Omega \subset B_{R/2}(0)$. For every function $\varphi \in C_c^{\infty}(\Omega)$, we define an extension $\tilde{\varphi} \colon B_R(0) \to \mathbb{R}$ by
\begin{align}
\tilde{\varphi}(x) &:= \varphi(x) &\text{ for } x \in \Omega, \\
\tilde{\varphi}(x) &:= 0 &\text{ otherwise.}
\end{align}
Then $\tilde{\varphi} \in C_{c}^{\infty}(B_R(0))$.
Let $\varphi, \psi \in C_c^{\infty}(\Omega)$. As above, we construct extensions $\tilde{\varphi}, \tilde{\psi} \in C_c^{\infty}(B_R(0))$. As the ball $B_R(0)$ has a smooth boundary, we are allowed to integrate by parts. So we obtain that
\begin{equation}
\int_{\Omega} \varphi \Delta \psi = \int_{B_R(0)} \tilde{\varphi} \Delta \tilde{\psi} = - \int_{B_R(0)} \nabla \tilde{\varphi} \cdot \nabla \tilde{\psi} = - \int_{\Omega} \nabla \varphi \cdot \nabla \psi.
\end{equation}
This proves the claim for bounded domains $\Omega$.
If $\Omega \subset \mathbb{R}^d$ is not bounded, then there is no ball $B_R(0) \subset \mathbb{R}^d$ such that $\Omega \subset B_{R/2}(0)$. However, our construction from above works! This is because for $\varphi \in C_c^{\infty}(\Omega)$ the support of $\varphi$ is a compact set; in particular, $\text{supp} \, \varphi \subset \mathbb{R}^d$ is bounded. So we find a ball $B_R(0) \subset \mathbb{R}^d$ such that $\text{supp} \, \varphi \subset B_{R/2}(0)$. Now, you can define the same extension $\tilde{\varphi} \colon B_R(0) \to \mathbb{R}$ as above. This proves the claim for unbounded domains $\Omega$.
Best Answer
If $M$ is $n$-dimensional Riemannian manifold and if $\Omega$ is a domain in $M$ with a compact closure, then assume that for a smooth function $f: M\rightarrow {\bf R}$, $f^{-1}(t)$ is hypersurface in $\Omega$. Then for $F: M\rightarrow {\bf R}$, $$ \int_{ \Omega} F d{\rm Vol} =\int_{(-\infty,\infty)} \int_{S_t} \frac{F}{|\nabla f|} d{\rm Vol}_{S_t} dt $$ where $$S_t:= f^{-1}(t) \cap \Omega $$
Proof : Consider a vector field $\frac{\nabla f}{|\nabla\ f|^2} $ whose integral curve is $\psi (t,x)$. Then $\frac{\partial }{\partial t} f\circ \psi(t,x) =1$ so that $f\circ\psi (t,x) - f\circ \psi(x,0) =\int_0^t 1=t$ Hence $$ \psi (t,S_{t_0}) =S_{t_0+t} $$
Note that for small $t$, volume between $S_{t_0},\ S_{t_0+t}$ in $\Omega$ is approximately $$\int_{S_{t_0}} {\rm length}\ \psi|_{[t_0,t_0+t]} d{\rm Vol}_{S_{t_0}} \sim \int_{S_{t_0}} \frac{t}{|\nabla f|} d{\rm Vol}_{S_{t_0}} $$ By Cavalieri's principle, we complete the proof.