= Solution
Identify $A$ with its image under $\alpha$. In a language naming every element of $B$, use the same names for elements of $A$ in the two sets of sentences
$$
\operatorname{Diag}(B)\cup\operatorname{ElDiag}(A).
$$
This union is consistent. To see finite satisfiability, conjoin the finitely many diagram facts about $B$ into a <quantifier-free formula> $\delta(\mathbf a,\mathbf b)$ and the finitely many elementary-diagram facts about $A$ into $\psi(\mathbf a)$, where the elements in $\mathbf b$ have names outside $A$. Since $B\models\exists\mathbf y\,\delta(\mathbf a,\mathbf y)$, the <1-embedding> property gives a witnessing tuple in $A$. Interpret the finitely many new names by that tuple. The resulting expansion of $A$ satisfies $\delta$ as well as $\psi$.
By <compactness theorem>, a model $C$ of this union exists. Its named copy of $B$ gives an embedding $\beta:B\to C$, while its named copy of $A$ is elementary. Consequently
$$
\boxed{\beta\alpha:A\longrightarrow C\text{ is elementary}.}
$$
The naming ensures that the two maps agree on the original copy of $A$.
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