Yes, that is a prime ideal.  If a product $g\cdot h$ 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.  Thus, the zero loci of $g$, resp. $h$ on $Z_{\text{sm}}$ are complex analytic subvarieties of a connected, complex manifold.  If neither of these complex analytic subvarieties equal all of $Z_{\text{sm}}$, then they are each nowhere dense, so that also the union is nowhere dense.  Thus, one of these, say $g$ vanishes identically on $Z_{\text{sm}}$.  Since $Z_{\text{sm}}$ is dense in $Z$ for the analytic topology, also $g$ vanishes identically on $Z$.  

Now you can use vanishing of cohomology of coherent analytic sheaves on Stein analytic spaces.  For the ideal $I$, define $\mathcal{I}\subset \mathcal{O}^{\text{an}}$ to be the image of the natural homomorphism, $$I\otimes_{\mathbb{C}[z_1,\dots,z_n]}\mathcal{O}^{\text{an}} \to \mathcal{O}^{\text{an}}.$$  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, $$0\to H^0(\mathbb{C}^n,\mathcal{K}) \to I\otimes_{\mathbb{C}[z_1,\dots,z_n]}H^0(\mathbb{C}^n,\mathcal{O}^{\text{hol}})\to H^0(\mathbb{C}^n,\mathcal{I}) \to 0.$$  Thus the element $g\in H^0(\mathbb{C}^n,\mathcal{I})$ is in the image of $I\otimes_{\mathbb{C}[z_1,\dots,z_n]} H^0(\mathbb{C}^n,\mathcal{O}^{\text{hol}}).$