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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 $\mathcal{U}$-ultraproducts \mathcal{A}$-ultraproducts are isomorphic iff$A \cong B$. 2 added 275 characters in body As you might expect, things are consistently much more interesting if$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$. 1 As you might expect, things are consistently much more interesting if$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}\$.