= Solution
Adopt the planar <Fourier transform> convention
$$
\widetilde g(\mathbf k)=\int e^{-i\mathbf k\cdot\mathbf x}g(\mathbf x)\,d^2x,\qquad
g(\mathbf x)=\int\frac{d^2k}{(2\pi)^2}e^{i\mathbf k\cdot\mathbf x}\widetilde g(\mathbf k).
$$
Extend the disk potential to three dimensions. The <Poisson equation for Newtonian gravity> becomes
$$
(\partial_z^2-k^2)\widetilde\phi(\mathbf k,z)=4\pi G\widetilde\Sigma(\mathbf k)\delta(z),\qquad k=|\mathbf k|.
$$
For $k>0$, decay away from an isolated disk and continuity across it give $\widetilde\phi=B(\mathbf k)e^{-k|z|}$. Integration through the plane yields $-2kB=4\pi G\widetilde\Sigma$, hence the <razor-thin disk Poisson kernel>
$$
\widetilde\phi(\mathbf k,0)=-\frac{2\pi G}{k}\widetilde\Sigma(\mathbf k).
$$
The given transform of $K(\mathbf x)=1/|\mathbf x|$ is $\widetilde K=2\pi/k$. The <convolution theorem> therefore gives the forward potential relation
$$
\boxed{\phi(\mathbf x)=-G\int\frac{\Sigma(\mathbf x')}{|\mathbf x-\mathbf x'|}\,d^2x'.}
$$
For the reverse relation, $\widetilde{\Delta_2\phi}=-k^2\widetilde\phi$, so
$$
\widetilde{K*\Delta_2\phi}
=\frac{2\pi}{k}(-k^2\widetilde\phi)
=4\pi^2G\widetilde\Sigma.
$$
Invert the transform to obtain the <thin-disk potential-density inversion>
$$
\boxed{\Sigma(\mathbf x)=\frac1{4\pi^2G}\int\frac{\partial_{x'}^2\phi(\mathbf x')+\partial_{y'}^2\phi(\mathbf x')}{|\mathbf x-\mathbf x'|}\,d^2x'.}
$$
An additive constant in $\phi$ is annihilated by $\Delta_2$. These formulas hold for suitable isolated finite disks, and in a distributional or regularized sense for less regular profiles. The uniform zero-wavenumber sheet needs a separate vertical solution; the infinite <Mestel disk> requires potential differences or an infrared regularization, since its absolute forward integral diverges.
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