Yes, that is a prime ideal. If a product $g\cdot h$ of holomorphic functions is contained in $I_{\text{hol}}$ then it vanishes on the zero scheme $Z$ of $I$. Since $Z$ is irreducible, the smooth locus $Z_{\text{sm}} = Z\setminus Z_{\text{sing}}$ is a connected, complex manifold. The zero loci of $g$, and $h$ on $Z_{\text{sm}}$ are complex analytic subvarieties of a connected, complex manifold. If neither of these complex analytic subvarieties equals all of $Z_{\text{sm}}$, then they are each nowhere dense. In that case, also the union is nowhere dense, contradicting the hypothesis that $g\cdot h$ vanishes on $Z$. Thus, one of the factors, say $g$, vanishes identically on $Z_{\text{sm}}$. Since $Z_{\text{sm}}$ is dense in $Z$ for the analytic topology, also $g$ vanishes on $Z$.
Now you can use vanishing of cohomology of coherent analytic sheaves on Stein analytic spaces. Fix a generating set $f_1,\dots,f_r$ of the ideal $I$. Define $\mathcal{I}\subset \mathcal{O}^{\text{an}}$ to be the image of the natural homomorphism, $$(\mathcal{O}^{\text{an}})^{\oplus r} \to \mathcal{O}^{\text{an}}, \ \ (u_1,\dots,u_r)\mapsto u_1f_1 + \dots + u_rf_r.$$ Define $\mathcal{K}$ to be the kernel of that map. Then $\mathcal{K}$ and $\mathcal{I}$ are coherent analytic sheaves. Since $H^1(\mathbb{C}^n,\mathcal{K})$ is zero, the induced long exact sequence of cohomology is a short exact sequence, $$0\to H^0(\mathbb{C}^n,\mathcal{K}) \to H^0(\mathbb{C}^n,\mathcal{O}^{\text{hol}})^{\oplus r} \to H^0(\mathbb{C}^n,\mathcal{I}) \to 0.$$ Thus the element $g\in H^0(\mathbb{C}^n,\mathcal{I})$ is in the image of $I\otimes_{\mathbb{C}[z_1,\dots,z_n]} H^0(\mathbb{C}^n,\mathcal{O}^{\text{hol}}).$