If $\mathcal A$ is a substructure of $\mathcal B$ and $\mathcal A$ is isomorphic to $\mathcal B$, then is it true that $\mathcal A$ is an elementary substructure of $\mathcal B$?
This question arose while I was trying to determine all the elementary substructure of $\mathbb Z$, viewed as a structure for the language $\mathcal L := \{ 0, +, - \}$ as an abelian group.
I have figured that any elementary substructure of $\mathbb Z$ must be of the form $a\mathbb Z$ for some $a \neq 0$. Now I am trying to show the converse. I have determined that $a\mathbb Z$ is isomorphic to $\mathbb Z$ for any $a \neq 0$. If the answer to my first line of this post is affirmative, then I will get my result.
We know that if $\mathcal A$ is an elementary substructure of $B$, then in particular $\mathcal A$ is equivalent to $\mathcal B$.
If the answer to my first line of this post is no, then is there a way to determine all the elementary substructure of $\mathbb Z$?
The answer to your main question is no, and in fact $\mathbb{Z}$ itself gives us a counterexample: the substructure $\mathbb{E}$ of even integers is clearly isomorphic to $\mathbb{Z}$ itself, but is not an elementary substructure. The failure of elementarity is a good exercise, so I've spoilered it below:
And this gives us, with a bit of care, a complete characterization of the elementary substructures of $\mathbb{Z}$:
In general, elementary substructurehood is an extremely strong condition. Of course it doesn't imply isomorphism-with-the-whole either (consider e.g. the rationals and the reals as linear orders), but most of the time the surprising part of the picture is how strong $\preccurlyeq$ is.