You may find this answer of mine of some interest. There, I explain a special case when one can prove that $I(M \times N) \cong I(M) \times I(N)$ with simple differential geometric arguments involving sectional curvatures. And then (in the edit) I sketch a proof of a much more general result using some holonomy techniques taken from Kobayashi & Nomizu. In a nutshell, this result says that if you have a product of some irreducible Riemannian manifolds and at most one flat Riemannian manifold, then the isometry group of the product is the product of the isometry groups of the factors plus permutations of isometric factors (no compactness or even completeness assumption required). One can also try to drop the irreducibility assumption, but then to say something interesting about the isometry group and how it relates to those of the factors, one needs to add the assumptions of completeness and simply connectedness (at least I don't see how to proceed otherwise). Here is the resulting statement:

**Proposition.** Let $M = M_1 \times \ldots \times M_k$ be a Riemannian product of complete simply connected Riemannian manifolds. We have an obvious embedding of Lie groups $i \colon I(M_1) \times \ldots \times I(M_k) \hookrightarrow I(M)$. The following are equivalent:

- $i$ is a local isomorphism, i. e. it gives an isomorphism on the identity components of the above groups: $I^0(M_1) \times \ldots \times I^0(M_k) \simeq I^0(M)$;
- $\dim I(M) = \dim I(M_1) + \cdots + \dim I(M_k)$;
- At most one of $M_i$'s has a nontrivial Euclidean de Rham factor.

Moreover, if these conditions are satisfied, then $i$ is an isomorphism if and only if there does not exist $i \ne j$ such that $M_i$ and $M_j$ share isometric de Rham factors. (If all $M_i$'s are irreducible, it means that no two of them are isometric.)

In order to prove this, just de Rham decompose each factor, stack all Euclidean subfactors together, and apply the second proposition from my answer linked above.