Edit 2 [After the updated question, one should probably read the comments below first before reading this]: A few things

(i) Intuition from 1+1 dimension can be misleading. Waves don't decay there. In the 1+3 case, solutions to the wave equation disperses. So for the equation give by $L = -c^{-2}\partial_t^2 + \triangle$, even when the wave speed is bigger than the speed of gravity, the solution is not incompatible with mere decay at spatial infinity. (Though of course, since the equation is no-longer Lorentz invariant, speaking of Spatial Infinity is somewhat of a red-herring: the conformal compactification of the space-time does not give a compatible compactification of the solution to the equation.) If to rule out such cases you need to also impose a rate of decay.

(ii) When I said local well-posedness of $L$, implicitly I mean on a suitable function space on space-like slices. For (strictly/symmetric/regularly) hyperbolic equations, you can of course study the characteristic cones and compare against the back ground null cone to have the decay automatically also satisfied for boosted slices.

(iii) As a trivial example, also note that in the $\partial_t v + Av$ formulation, you can take the Fourier transform to get the solution to be $\hat{v}(t,\xi) = e^{-A(\xi)t}\hat{v}_0(\xi)$. If the matrix $A(\xi)$ has only eigenvalues with negative real parts, then Schwarz data will lead to Schwarz solutions, and lead to non-uniqueness of the solutions. This is sort of the explicit version of the Hille-Yoshida type theorems.

I am somewhat doubtful that the question as posed as any sort of reasonable answer. (Also, I don't really see how the Lorentzian metric even enter into the problem.)

(a) ANY hyperbolic PDE in (1+3) dimensions will, by definition, not have unique solution for your problem.

(b) Even elliptic PDEs are not guaranteed to have unique solutions, even for nice ones that come from minimization of a strictly convex energy functional: Euclidean space simply has too large a symmetry group, and if the equation itself is invariant under translations, any finite translate of a solution that decays at spatial infinity is another solution.

(c) If you restrict yourself to linear scalar PDEs $L[u] = 0$, by linearity, $u \equiv 0$ is necessarily a solution that satisfy your conditions. So your question as posed reduces to "which linear operator admits only the trivial solution". For constant coefficient $L$'s you can simply take the Fourier transform of both sides and study the zero-set of the PDO.

If you clarify your motivation for considering the question your are asking, maybe more reasonable answers can be given.

I am somewhat doubtful that the question as posed as any sort of reasonable answer. (Also, I don't really see how the Lorentzian metric even enter into the problem.)

(a) ANY hyperbolic PDE in (1+3) dimensions will, by definition, not have unique solution for your problem.

(b) Even elliptic PDEs are not guaranteed to have unique solutions, even for nice ones that come from minimization of a strictly convex energy functional: Euclidean space simply has too large a symmetry group, and if the equation itself is invariant under translations, any finite translate of a solution that decays at spatial infinity is another solution.

(c) If you restrict yourself to linear scalar PDEs $L[u] = 0$, by linearity, $u \equiv 0$ is necessarily a solution that satisfy your conditions. So your question as posed reduces to "which linear operator admits only the trivial solution". For constant coefficient $L$'s you can simply take the Fourier transform of both sides and study the zero-set of the PDO.

If you clarify your motivation for considering the question your are asking, maybe more reasonable answers can be given.

Edit: To say a bit more about the Fourier transform method. For example, if you assume that your solution is in Schwarz space on spatial slices, and assume the the PDO $L$ does not have $\partial_t$ terms, then you can ignore the time component. Restricting to the spatial slice and taking Fourier transform, you see that $P(\xi)\hat{u}(\xi) = 0$ where $P$ is the symbol for $L$, and is a polynomial in $\xi$. Using that the Fourier transform is injective on $\mathcal{S}$, you immediately have that $\hat{u}$ can only be supported on the zero set of $P$, but as $P$ is a polynomial, its zero set has vanishing measure unless $P\equiv 0$. Thus you see that in Schwarz space any non-trivial constant coefficient $L[u] = 0$ only has trivial solutions. Now, it is well known that the smoothness of $\hat{u}$ is related to the decay properties of $u$. By considering restriction theorems/Strichartz type estimates, there can be solutions for certain PDOs in $L^p$. (As an example, let the space-dimension be 4 [only because I remember the Lebesgue exponent explicitly in this case], Then the PDO on $(t,w,x,y,z)$ given by $-\partial_w^2 + \partial_x^2 + \partial_y^2 + \partial_z^2$ admits infinitely many solutions in $L^4(\mathbb{R}^4)$, with Fourier transform supported on the set $\hat{w}^2 = \hat{x}^2 + \hat{y}^2 + \hat{z}^2$. You can even ask that the solutions be smooth: they just cannot decay too fast in the $w$ direction.)