 Solution

If $F$ is any other fundamental solution, then $H=F-E$ satisfies ${\partial }_{\bar{z}}H=0$ as a distribution. Applying $4{\partial }_z$ gives $\Delta H=0$. Elliptic regularity for the Laplacian implies that the initially arbitrary distribution $H$ is a smooth function. Its equation is then the ordinary Cauchy-Riemann equation, so $H$ is an entire function.

For a direct justification of the needed regularity, choose a radial mollifier ${\eta }_r$. Locally, $H\ast {\eta }_{\epsilon }$ is a smooth harmonic function. On a ball with sufficient room for both radii, the mean value property gives $(H\ast {\eta }_{\epsilon })\ast {\eta }_r=H\ast {\eta }_{\epsilon }$. Passing to the weak convergence of distributions limit $\epsilon \to 0$ yields $H\ast {\eta }_r=H$ on a smaller ball. The left side is smooth, proving the claim locally. This establishes the distributional Cauchy-Riemann lemma without assuming in advance that $H$ is a locally integrable function.

Conversely every entire function $h$ satisfies ${\partial }_{\bar{z}}h=0$ and can be added to $E$. Hence all fundamental solutions areNo growth condition at infinity was imposed, so $h$ need not be a polynomial.