Is the extension (dual restriction) operator on any smooth hypersurface a solution to some PDE? We know that the extension operator on paraboloids $\widehat{fd\sigma}(t,x)=\int_\mathbb{R}^nf(\xi)e^{i(t|\xi|^2+x\cdot\xi)}d\xi$ is a solution to the homogeneous Schrodinger equation with initial data $f$; that on cones (change $|\xi|^2$ to $|\xi|$) a solution to the wave equation with the same initial data; and on (part of) a spheres (change $|\xi|^2$ to $\sqrt{1-|\xi|^2}$) a solution to the Hamiltonian equation.
Is every extension operator on some smooth hypersurface (say, quadratic surfaces like hyperbolic hyperboloids, hyperbolic paraboloids, or more general surfaces that are not graphs of any quadratic or even polynomial functions) a solution to some PDE?
 A: Your question was basically answered by David Roberts in the comments, but let me write a few more words.
Given a constant coefficient linear differential operator of degree $N$
$$ L = \sum_{|\alpha| \leq N} c_\alpha \partial^\alpha $$
(here I use multi-index notation for $\alpha$), we can formally take the Fourier transform of the equation $L u = 0$ to get
$$ \widehat{(Lu)}(\xi) = \sum_{|\alpha| \leq N} c_\alpha i^{|\alpha|} \xi^\alpha \hat{u}(\xi) = P(\xi) \hat{u}(\xi) = 0 $$
Here $P(\xi)$ is a polynomial function. So within the space of tempered distributions, you have that solving $Lu = 0$ is the same as finding a distribution satisfying $P(\xi)\hat{u}(\xi) = 0$. This requires that the support of $\hat{u}$ be on the 0 set of $P$.
So this gives you the general correspondence between Fourier extensions on algebraic varieties and solutions to PDEs.

However, this correspondence is not one-to-one in general.
Take the simplest case of an ODE on $\mathbb{R}$.
Let $L_1 = \frac{d}{dx}$ and $L_2 = \frac{d^2}{dx^2}$. Their corresponding polynomials are $i \xi$ and $- \xi^2$. Both have the same zero set.
But in general the solutions to $L_1 u = 0$ are just the constant distributions (whose Fourier transform are multiples of $\delta_0$). But solutions to $L_2 u = 0$ include all linear functions (Fourier transform is a linear combination of $\delta_0$ and $\delta_0'$).
This shows that not all constant coefficient linear PDEs have their solutions expressible, in Fourier space, as $\hat{f}$ times the surface measure of some surface.

Finally, your question also asked about smooth hypersurfaces which may not be algebraic varieties. Let $\phi$ be a defining function of your hypersurface, then every surface measure $f d\sigma$ has the property that $\phi(\xi) \cdot f d \sigma = 0$, and so is (by definition) the Fourier transform of a distributional solution of the pseudo-differential equation $\phi(D)u = 0$.
In general the pseudodifferential operator $\phi(D)$ may fail to be a local operator, and so the corresponding equation may fail to be a PDE.
