We can get more examples from measurable cardinals and measurable limits of measurable cardinals.
For every set of ultrafilters $M$, there is a structure $\mathcal{A}$ such that if $\mathcal{U},\mathcal{V}\in M$, then there is an elementary embedding from $\mathcal{A}^\mathcal{U}$ to $\mathcal{A}^\mathcal{V}$ if and only if $\mathcal{U}\leq_{RK}\mathcal{V}$ in the Rudin-Kiesler ordering (for example, $\mathcal{A}$ could a structure with cardinality greater than the underlying set of each ultrafilter on $M$ and where every operation is a fundamental operation and every relation is a fundamental relation). I believe that this correspondence between elementary embeddings and the Rudin-Kiesler ordering can be found in Andreas Blass' dissertation. I think you can also let $\mathcal{A}=V_\alpha$ for large enough $\alpha$.
Suppose that $I$ is a set. Suppose that $\mathcal{U}$ is a non-principal ultrafilter on $I$, and $\mathcal{U}_i$ is an ultrafilter on some set $X_i$ for $i\in I$. For simplicity, suppose that $X_i\cap X_j=\emptyset$ for $i\neq j$. Let $\mathcal{V}$ be the ultrafilter on $\bigcup_{i\in I}X_i$ where $R\in\mathcal{V}$ precisely when $\{i\in I\mid R\cap X_i\in\mathcal{U}_i\}\in\mathcal{U}$. We observe that $\prod_{i\in I}\mathcal{A}^{\mathcal{U}_i}/\mathcal{U}\simeq\mathcal{A}^{\mathcal{V}}$ for all structures $\mathcal{A}$. We shall now produce examples where $\mathcal{U}_i\not\leq_{RK}\mathcal{V}$ for $i\in I$.
(1) Suppose that $\mu$ is a measurable limit of measurable cardinals. Let $I$ be a set of measurable cardinals below $\mu$ with $|I|=\mu$. Suppose that the ultrafilter $\mathcal{U}$ is $\mu$-complete and $\mathcal{U}_\alpha$ is $\alpha$-complete but not $\mu$-complete whenever $\alpha\in I$. Then the ultrafilter $\mathcal{V}$ is $\mu$-complete. However, if $\mathcal{W}\leq_{RK}\mathcal{V}$, then $\mathcal{W}$ must also be $\mu$-complete, so $\mathcal{U}_\alpha\not\leq_{RK}\mathcal{V}$ for $\alpha\in I$.
(2). I claim that if $\mu$ is a measurable cardinal, $|I|<\mu$, and each $\mathcal{U}_i$ is $\mu$-complete, and $\mathcal{W}$ is $\mu$-complete, then whenever $\mathcal{W}\leq_{RK}\mathcal{V}$, there is some $Q^\sharp\in \mathcal{U}$ where $\mathcal{W}\leq_{RK}\mathcal{U}_i$ for $i\in Q^\sharp$. Suppose that $\mathcal{W}$ is supported on a set $X$ and $f:\bigcup_{i\in I}X_i\rightarrow X$ is a function such that if $R\subseteq X$, then $R\in\mathcal{W}$ if and only if $f^{-1}[R]\in\mathcal{V}$. If $R\subseteq X$, then let $R^\sharp=\{i\in I:f^{-1}[R]\cap X_i\in\mathcal{U}_i\}$. Observe that if $R\in\mathcal{W}$, then $R^\sharp\in\mathcal{U}$ and the mapping $R\mapsto R^\sharp$ is a Boolean algebra homomorphism from $P(X)$ to $P(I)$. Let $\mathbf{T}=\{R^\sharp:R\in\mathcal{W}\}$. For each $S\in\mathbf{T}$, there is some $S_\sharp\in\mathcal{W}$ with $(S_\sharp)^\sharp=S$. Therefore, by $\mu$-completeness and since $2^{|I|}<\mu$, we know that $\bigcap_{S\in\mathbf{T}}S_\sharp\in\mathcal{W}$. Therefore, if we set
$Q=\bigcap_{S\in\mathbf{T}}S_\sharp$, then $Q^\sharp$ is the smallest element in $\mathbf{T}$. Now let $\mathcal{W}_1=\mathcal{W}\cap P(Q)$. Then $R^\sharp=Q^\sharp$ for $R\in\mathcal{W}_1$ and $R^\sharp=\emptyset$ for $R\in P(Q)\setminus\mathcal{W}_1$. If $i\in Q^\sharp$, then $f^{-1}[R]\cap X_i\in\mathcal{U}_i$ for $R\in\mathcal{W}_1$. Therefore, $\mathcal{W}\leq_{RK}\mathcal{U}_i$ for $i\in I$.
In particular, if $\mathcal{U}_j\leq_{RK}\mathcal{V}$, then there is some $Q^\sharp\in\mathcal{U}$ where $\mathcal{U}_j\leq_{RK}\mathcal{U}_i$ for $i\in Q^\sharp$. But there are a few ways to ensure that we cannot have $\mathcal{U}_j\leq_{RK}\mathcal{V}$ including the following:
i. Suppose that $\mu_i$ is a measurable cardinal with $\mu_i>|I|$ for $i\in I$, and $\mathcal{U}_i$ is a normal ultrafilter on $\mu_i$ for $i\in I$. Suppose furthermore that $\mathcal{U}_i\neq\mathcal{U}_j$ for $i\neq j$. Then since normal ultrafilters are Rudin-Keisler minimal, if $\mathcal{U}_j\leq_{RK}\mathcal{U}_i$, then either $\mathcal{U}_j$ is principal or $\mathcal{U}_j=_{RK}\mathcal{U}_i$ which contradicts our hypotheses.
ii. Suppose that $I$ is a well-ordered set and $\mu_i$ is a measurable cardinal with $\mu_i>|I|$ for $i\in I$ and that $\mu_i<\mu_j$ whenever $i<j$. Suppose furthermore that if $i<j$, then $\mathcal{U}_i$ is $\mu_i$-complete but where $|X_i|<\mu_j$. Then whenever $\mathcal{V}\leq_{RK}\mathcal{U}_i$ and $i<j$, the ultrafilter $\mathcal{V}$ is $\mu_i$-complete, but not $\mu_j$-complete, so if
$\mathcal{U}_j\leq_{RK}\mathcal{U}_i$, then $i=j$.