 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 sentencesThis 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. ConsequentlyThe naming ensures that the two maps agree on the original copy of $A$.