An obvious obstruction to the proposed topological characterization is the following: Endow $H(D)$ with the discrete topology, call it $\tau$. Then the space $\langle H(D) , \tau\rangle$ has the following properties: $f \mapsto f(z)$ is continuous for every $z$, $\langle H(D), \tau\rangle$ is a topological ring (as in, point-wise addition and point-wise multiplication of functions are continuous), and $\langle H(D), \tau\rangle$ is metrizable. The obvious down-side to this topology is that it is not separable and not natural in the same way but it does offer a counter-example to the statement
If $\tau$ is a (Hausdorff) topology on $H(D)$ for which point-evaluation is continuous, the group and ring operations are continuous, and $H(D)$ is metrizable, then $\tau$ must be the compact-open topology.
Edit: If we are only interested in separable and complete metrizable topologies, we can say a little more.
In https://arxiv.org/abs/2101.07386, it is shown that,
if $H(D)$ where $D$ is an open subset of $\mathbb C$ has a Polish topology for which the ring operations are continuous, then that topology must be the topology of uniform convergence on compact subsets.
A similar result about domains of $\mathbb C^n$ for $n \geq 2$ is shown where the necessary assumption is something a little less than being a Gleason domain of holomorphy (we just need the maximal ideals to be generated in the correct fashion for a dense subset of the domain).