= Solution
For the rotationally invariant free-space <fundamental solution of the Laplace equation>, put $s=\sqrt{(x-x')^2+(y-y')^2}$. Away from the source, the <radial Laplacian> gives $(sG')'=0$, hence $G=A\log s^2+C$. Integrate its <Poisson equation> over a disk of radius $r$ about the source. The <divergence theorem> gives $2\pi r(2A/r)=-1/\epsilon_0$, so
$$
\boxed{G=-\frac{1}{4\pi\epsilon_0}\log\big((x-x')^2+(y-y')^2\big)+C.}
$$
A fixed reference length can be inserted inside the <logarithm> and absorbed into $C$. The resulting <electric field> decays like $1/s$.
Strictly, the stated pointwise decay as $y\to\pm\infty$ alone does not justify “most general”: one can also add the <harmonic function> $\operatorname{Re}e^{(x+iy)^2}$, whose <gradient> has that decay for every fixed $x$. This is the same growth issue as a <zero-boundary harmonic function with pointwise vertical decay>. The displayed answer is the usual isotropic free-space choice; requiring the field to vanish uniformly in every direction at spatial infinity removes these extra harmonic additions. An additive constant remains undetermined.
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