Let the terminology and notation of the book Model Theory (1990) by Chen-Chung Chang and H. Jerome Keisler be in force.
Question
Is there a quintuple $\langle\mathcal L,A,\mathcal I,D,E\rangle$ satisfying Conditions (1), (2), (3) and (4) below?
(1) $\mathcal L$ is a language,
(2) $\langle A,\mathcal I\rangle$ is a model for $\mathcal L$,
(3) $D$ and $E$ are non-principal ultrafilters over $\omega$,
(4) we have $$ \prod_D\langle A,\mathcal I\rangle\not\cong\prod_E\langle A,\mathcal I\rangle. $$
The article [1] implies that a negative answer to the above question cannot be proved in ZFC.
Corollary 6.1.2 in Chang and Keisler's book implies that, if $\langle\mathcal L,A,\mathcal I,D,E\rangle$ satisfies (1), (2), (3) and (4), then we have $|\mathcal L|>\aleph_0$ or $|A|>2^{\aleph_0}$.
Note that, in the above setting, the models $\prod_D\langle A,\mathcal I\rangle$ and $\prod_E\langle A,\mathcal I\rangle$ have same cardinal, as follows from the Frayne-Morel-Scott Theorem, a proof of which can be found in this text of Keith Kearnes.
Here is a related question.
[1] Roitman, Judy. "Non-isomorphic hyper-real fields from non-isomorphic ultrapowers." Mathematische Zeitschrift 181, no. 1 (1982): 93-96, link.
The answer is yes.
Consider the structure $M$ with the universe $\omega$ with predicates for all subsets of $\omega$.
Given any $N\equiv M$ and $a\in N$, the set $\{A\subseteq \omega| N\models a\in A \}$ is an ultrafilter on $\omega$. Given an $N$, denote by $U(N)$ the set of all these ultrafilters.
It is easy to see that for any countable ultrapower $N$ of $M$, the cardinality of $U(N)$ is clearly at most $\lvert N\rvert \leq 2^{\aleph_0}$. It is also clear that $U(N)$ is an isomorphism invariant.
Now, notice that if $\mathcal U$ is any ultrafilter, then $\mathcal U\in U(M^\mathcal U)$, as witnessed by $[n]_{n\in \omega}$. Since there are $2^{2^{\aleph_0}}$ ultrafilters on $\omega$, it follows that there are $2^{2^{\aleph_0}}$ isomorphism classes of countable ultrapowers of $M$.