This $\Longrightarrow$ implies the desired $\Longleftrightarrow$. It would indeed make more sense to define an elementary embedding $f$ to be a map $f : A \mapsto B$ satisfying $\forall \phi \ \forall \bar{a} \in A, \, A\models\phi[\bar{a}]\Longleftrightarrow B\models\phi[f\bar{a}]$ (Btw notice that this implies that $f$ is injective). Then, you can make the following observation :
Claim Assume a map $f : A \mapsto B$ satisfies $\forall \phi \ \forall \bar{a} \in A, \, A\models\phi[\bar{a}]\Longrightarrow B\models\phi[f\bar{a}]$. Then $f$ is an elementary embedding.
Let $f$ be as in claim, we show that $f$ is an elementary embedding :
All what remains to do is to show that if $B \models \phi[f\bar{a}]$ for some $\phi, \bar{a}$, then $A\models \phi[\bar{a}]$. Assume for contradiction that this does not hold, then $A \models \neg \phi[\bar{a}]$ and by assumption we get $B \models \neg \phi[f\bar{a}]$, a contradiction to our assumption that $B\models \phi[f\bar{a}]$ !