= Solution
A <model-theoretic ultralimit> is the direct limit of a chain of successive <ultrapowers> along their diagonal elementary maps. It is not the topological notion of a limit along an <ultrafilter>. We first prove the precise relative embedding lemma needed to synchronize two such limits.
If $A\prec B$, then $B$ embeds elementarily into an <ultrapower> of $A$, with the embedding restricting on $A$ to its diagonal map. To prove this <Frayne embedding lemma>, use the elementary diagram of $B$ with the elements of $A$ treated as named parameters. Every finite fragment $\Delta$ of that diagram has a realization in $A$: replace its finitely many names from $B\setminus A$ by variables and existentially quantify their conjunction. It is true in $B$, and hence true in $A$ by elementarity. Choose these simultaneous realizations for each finite fragment, keeping every parameter from $A$ interpreted as itself; give unmentioned names an arbitrary default in the nonempty $A$.
On the index set of finite fragments, extend the filter of cones $\{\Delta:\Delta\supseteq\Gamma\}$ to an <ultrafilter> $U$. For each $b\in B$, let $g_b(\Delta)$ be its value in the chosen coordinate realization. Every diagram sentence is satisfied on its cone, so the <Łoś theorem> makes $b\mapsto[g_b]$ an <elementary embedding> $B\to A^I/U$. Inequalities in the diagram ensure injectivity. If $b\in A$, all its coordinates equal $b$, giving exactly the diagonal map.
There is also the initial version: if merely $A\equiv B$, then $A$ embeds elementarily into an <ultrapower> of $B$. A finite fragment of the elementary diagram of $A$, with all its names replaced by variables, yields an existential sentence true in $A$ and hence in $B$. The identical finite-fragment and cone construction therefore proves this version without already assuming an embedding.
Start with $A_0=A$. Use the initial version to place $A_0$ elementarily in an <ultrapower> $B_0$ of $B$. Relabel embedded copies so we can regard the elementary maps as inclusions. Apply the relative version to $A_0\prec B_0$ to obtain an <ultrapower> $A_1$ of $A_0$ containing $B_0$, and whose inclusion of $A_0$ is its diagonal map. Next apply it to $B_0\prec A_1$ to obtain an <ultrapower> $B_1$ of $B_0$ containing $A_1$, again fixing the earlier diagonal inclusion. Continuing gives an elementary chain
$$
A_0\prec B_0\prec A_1\prec B_1\prec A_2\prec B_2\prec\cdots.
$$
The common union $C$ is elementary over every stage by the <elementary chain theorem>, and
$$
C=\bigcup_{n<\omega}A_n=\bigcup_{n<\omega}B_n.
$$
The first presentation is a <model-theoretic ultralimit> of $A$. The second, preceded by the initial diagonal embedding of $B$ into $B_0$, is a <model-theoretic ultralimit> of $B$. Each consecutive same-letter inclusion is precisely the corresponding <ultrapower> diagonal map, not just an unspecified <elementary embedding>. Thus the two limits are identified with the same structure $C$, proving
$$
\boxed{A\equiv B\ \Longrightarrow\ A\text{ and }B\text{ have isomorphic model-theoretic ultralimits}.}
$$
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