As you might expect, things are consistently much more interesting if $CH$ fails. This has been explored by Shelah in a fascinating series of papers "Vive la difference I - III". For example, it is consistent that there is a nonprincipal ultrafilter $\mathcal{U}$ on $\omega$ such that if $(R_{n})$ and $(S_{n})$ are sequences of discrete rank 1 valuation rings having countable residue fields, then any isomorphism $\varphi: \prod_{\mathcal{U}}R_{n} \to \prod_{\mathcal{U}}S_{n}$ is an ultraproduct of isomorphisms $f_{n}: R_{n} \to S_{n}$. In particular, $\mathcal{U}$-almost all $R_{n}$ are isomorphic to the corresponding $S_{n}$ and so the Ax-Kochen isomorphism theorem doesn't hold with respect to $\mathcal{U}$.

If you are only interested in ultraproducts of fixed structures $A$, $B$, then I should mention that if $A$ and $B$ are countable structures which satisfy the strong independence property, then the corresponding $\mathcal{U}$-ultraproducts are isomorphic iff $A \cong B$.

As you might expect, things are consistently much more interesting if $CH$ fails. This has been explored by Shelah in a fascinating series of papers "Vive la difference I - III". For example, it is consistent that there is a nonprincipal ultrafilter $\mathcal{U}$ on $\omega$ such that if $(R_{n})$ and $(S_{n})$ are sequences of discrete rank 1 valuation rings having countable residue fields, then any isomorphism $\varphi: \prod_{\mathcal{U}}R_{n} \to \prod_{\mathcal{U}}S_{n}$ is an ultraproduct of isomorphisms $f_{n}: R_{n} \to S_{n}$. In particular, $\mathcal{U}$-almost all $R_{n}$ are isomorphic to the corresponding $S_{n}$ and so the Ax-Kochen isomorphism theorem doesn't hold with respect to $\mathcal{U}$.

If you are only interested in ultraproducts of fixed structures $A$, $B$, then I should mention that it is also consistent that there exists an ultrafilter $\mathcal{A}$ on $\omega$ such that if $A$ and $B$ are countable structures which satisfy the strong independence property, then the corresponding $\mathcal{A}$-ultraproducts are isomorphic iff $A \cong B$.

As you might expect, things are consistently much more interesting if $CH$ fails. This has been explored by Shelah in a fascinating series of papers "Vive la difference I - III". For example, it is consistent that there is a nonprincipal ultrafilter $\mathcal{U}$ on $\omega$ such that if $(R_{n})$ and $(S_{n})$ are sequences of discrete rank 1 valuation rings having countable residue fields, then any isomorphism $\varphi: \prod_{\mathcal{U}}R_{n} \to \prod_{\mathcal{U}}S_{n}$ is an ultraproduct of isomorphisms $f_{n}: R_{n} \to S_{n}$. In particular, $\mathcal{U}$-almost all $R_{n}$ are isomorphic to the corresponding $S_{n}$ and so the Ax-Kochen isomorphism theorem doesn't hold with respect to $\mathcal{U}$.

As you might expect, things are consistently much more interesting if $CH$ fails. This has been explored by Shelah in a fascinating series of papers "Vive la difference I - III". For example, it is consistent that there is a nonprincipal ultrafilter $\mathcal{U}$ on $\omega$ such that if $(R_{n})$ and $(S_{n})$ are sequences of discrete rank 1 valuation rings having countable residue fields, then any isomorphism $\varphi: \prod_{\mathcal{U}}R_{n} \to \prod_{\mathcal{U}}S_{n}$ is an ultraproduct of isomorphisms $f_{n}: R_{n} \to S_{n}$. In particular, $\mathcal{U}$-almost all $R_{n}$ are isomorphic to the corresponding $S_{n}$ and so the Ax-Kochen isomorphism theorem doesn't hold with respect to $\mathcal{U}$.

If you are only interested in ultraproducts of fixed structures $A$, $B$, then I should mention that if $A$ and $B$ are countable structures which satisfy the strong independence property, then the corresponding $\mathcal{U}$-ultraproducts are isomorphic iff $A \cong B$.