Yes, $\mathscr{O}(\mathbb{C})$ is a GCD domain, and as such it is integrally closed.
You can also see it directly, if $f \in \mathscr{M}(\mathbb{C})$ satisfies
$$\sum_{k=0}^n a_k(z) f(z)^k \equiv 0$$
with $a_k \in \mathscr{O}(\mathbb{C})$ and $a_n = 1$, if it had a pole of order $m$ in $z_0$, you would have
$$1 + \sum_{k=0}^{n-1} a_k(z)f(z)^{k-n} \equiv 0$$
in a punctured neighbourhood of $z_0$, but
$$\lim_{z\to z_0} \sum_{k=0}^{n-1} a_k(z)f(z)^{k-n} = 0$$
since the $a_k$ are bounded.