Let $\mathfrak A,\mathfrak A^*$ be $\mathcal L$-structures and $\mathfrak A \preceq \mathfrak A^*$. That implies forall n-ary formula $\varphi(\bar{v})$ in $\mathcal L$ and $\bar{a} \in \mathfrak A^n$
$$\models_{\mathfrak A}\varphi[\bar{a}] \iff \models_{\mathfrak A^*}\varphi[\bar{a}]$$
Therefore forall $\mathcal L$-sentence $\phi$
$$\models_{\mathfrak A}\phi \iff \models_{\mathfrak A^*}\phi$$
,which implies $\mathfrak A \equiv \mathfrak A^*$
But I haven't found this result in textbook, so I'm not sure.
Best Answer
(I realise that it was answered in the comments, but I'm posting the answer so as to keep the question from staying in the unanswered pool.)
This is, of course, true, an $\mathcal L$-sentence without parameters is an $\mathcal L$-sentence with parameters, that happens not to use any parameters, so elementary extension is a stronger condition. :)
To put it differently, $M\preceq N$ is equivalent to saying that $M$ is a substructure of $N$ and that $(M,m)_{m\in M}\equiv (N,m)_{m\in M}$, which is certainly stronger than mere $M\equiv N$ (to see that the converse does not hold, consider, for instance, $M=(2{\bf Z},+)$, $N=({\bf Z},+)$ -- $M$ is a substructure of $N$ and is e.e. (even isomorphic!) to it, but is still not an elementary substructure).