For any first order theory $T$, let's say that $M$ is an ideal existential model of $T$ if and only if, for every formula $\varphi(x)$ in the language of $T$ in which only the symbol "$x$" appears free, we have: $$M \models \exists x (\varphi(x)) \text{ if and only if }T\vdash \exists x (\varphi(x)). $$ In English, there exists $x$ fulfilling $\varphi(x)$ in the domain of the model $M$ of $T$ if and only if $T$ proves the existence of $x$ for which $\varphi(x)$ holds.
Question: Do ideal existential models exist? And if so, then can they exist for [A]ny first order theory? If only [S]ome, then which theories can have such models?
A theory $T$ has an ideal existential model if and only if $T$ is complete (and if $T$ is complete, then every model is ideal existential!).
Indeed, if $T$ is complete, then for any sentence $\psi$ and any $M\models T$, we have $M\models \psi$ if and only if $T\vdash \psi$.
Conversely, suppose $M$ is an ideal existential model of $T$. Note that for any sentence $\psi$, we can let $x$ be a variable which doesn't occur in $\psi$, and let $\widehat{\psi}(x)$ be the formula $(\psi\land (x = x))$. Then $\exists x\, \widehat{\psi}(x)$ is logically equivalent to $\psi$.
Now if $M\models \psi$, then $M\models \exists x\, \widehat{\psi}(x)$, so $T\vdash \exists x\, \widehat{\psi}(x)$, so $T\vdash \psi$. Since either $M\models \psi$ or $M\models \lnot \psi$, we have $T\vdash \psi$ or $T\vdash \lnot \psi$. Hence $T$ is complete.
(In this answer, I have ignored empty models - if your version of first-order logic allows empty models, then some details above have to be adjusted slightly, since $\psi$ and $\exists x\, \widehat{\psi}(x)$ are only logically equivalent in non-empty models. See Eric Wofsey's comment below. )