Even adding one set can break metrizability, if that set is not $F_\sigma$.

Let $\tau'$ be generated by $\tau$ and $A$, where $A$ is not $F_\sigma$ with respect to $\tau$. (For instance, by the Baire category theorem, $A = (\mathbb{Q} \times \mathbb{Q})^c$ would do.) Now if $\tau'$ is metrizable, then the open set $A$ must be $F_\sigma$ with respect to $\tau'$; indeed, we would have $A = \bigcup_{n=1}^\infty \bigcap_{x \in A^c} (B'(x, 1/n)^c)$. But I claim this is not so.

It's easy to verify that every open set $U'$ in $\tau'$ may be written as $U' = U \cup (A \cap V)$ where $U, V \in \tau$. (Check that the collection of all such sets is a topology which contains $\tau$ and $A$.) Now if $A$ is $F_\sigma$ in $\tau'$, then $A^c$ is $G_\delta$ in $\tau'$. So $A^c = \bigcap_{n=1}^\infty U_n'$ where $U_n' = U_n \cup (A \cap V_n) \in \tau'$, with $U_n, V_n \in \tau$. But every $U_n'$ must contain $A^c$, which means that $U_n' = U_n \cup V_n \in \tau$. So we conclude that $A^c$ is $G_\delta$ in $\tau$, a contradiction.