I'm trying to gain a better intuitive understanding behind the relationship of substructures and elementary substructures, and I've come to a contradiction that is very obviously false, so there must be some flaw in my understanding somewhere. Below is my thought process, please correct anything I say that is wrong.
From my understanding, if $\mathcal{A}$ is a substructure of $\mathcal{B}$ then $\mathcal{A}, \mathcal{B}$ agree on all quantifier-free atomic formulas when evaluated on $A$. That is, the structures say the same things are true for any individual elements of $A$, but not when talking about sweeping claims concerning the whole domain (i.e quantifiers). This is a somewhat strong requirement, as the structures essentially have to be the same aside from a differing domain, i.e when viewed only on $A$ they are essentially the same (is this true?)
From the definition on Wikipedia, my understanding is that elementary substructures must agree on all sentences. So to go from a substructure to elementary substructure, basically all we are adding is the agreement over quantifiers. In particular, you have to add all of the elements of $B$ that satisfy the $\exists$ claims that can be made for $\mathcal{B}$. That is, you have to add any definable subset of $B$. This is because $\mathcal{A}$ now satisfies all $\exists x\phi(x)$, so any $x$ that can be defined by some formula must be in $A$, and $\mathcal{A},\mathcal{B}$ agree about everything that can be said for elements in these definable sets. Now $\mathcal{A},\mathcal{B}$ are basically the same model except that their domains only differ by some non-definable set.
But so when exactly can they differ? The statement would have to be a formula with free variables, say $\psi(x)$, since they agree on all sentences. Suppose the structures disagree on this formula. If $\mathcal{B}\vDash \psi[b]$, then by definition there's some $b\in B$ such that $\mathcal{B}\vDash\exists x\psi(x)$. But $\exists x\psi(x)$ is a sentence, so $\mathcal{A}$ satisfies it, hence $\mathcal{A}\vDash\exists x\psi(x)$. What i'm assuming is the problem here is that for the structures to disagree on $\psi[b]$, we would need $b\in B\setminus A$, meaning $b$ is not definable. So where exactly in the above steps does the non-definability of $b$ break things? Does it not make sense to even say $\psi[b]$? Or is it in the next step, where we say $\mathcal{B}\vDash\exists x\psi(x)$? If so, I thought this was just the definition of satisfaction for existential sentences. I'm clearly very gravely misunderstanding something, so any help would be appreciated.