Let $\mathscr{L}'= \mathscr{L}\cup \mathscr{C}$ be an extension of the language $\mathscr{L}$ with a new infinite set of constants $\mathscr{C}$, and $T$ be an $\mathscr{L}$ theory. I wish to show that if we write $[\varphi]$ for the set of all formulas equivalent to $\varphi$ modulo $T$ (That is the set of all $\psi$, such that $T\models \varphi \longleftrightarrow \psi$) and if we view the set of all such $[\varphi]$ as a Boolean algebra then, for any $\mathscr{L}'$ formula with one free variable $\varphi(v)$:
$$[\exists v \varphi(v) ]= \bigvee_{c \in \mathscr{C}}[\varphi(c)]$$
From $T\models \varphi(c) \longrightarrow \exists v \varphi (v)$ we have $[\exists v \varphi(v) ] \geq \bigvee_{c \in \mathscr{C}}[\varphi(c)]$. How to prove the converse inequality?
We want to show that $[\exists v\,\varphi(v)]$ is the least upper bound of $\{[\varphi(c)]\mid c\in C\}$. You've already noted that for all $c\in C$, $[\varphi(c)]\leq [\exists v\,\varphi(v)]$, so it suffices to show that if $\psi$ is an $L'$-formula such that $[\varphi(c)]\leq [\psi]$ for all $c\in C$, then $[\exists v\,\varphi(v)]\leq [\psi]$.
Indeed, note that only finitely many of the constants in $C$ appear in the formula $\psi$. Let $d$ be a constant in $C$ not appearing in $\psi$. Then we have $T\models \varphi(d)\rightarrow \psi$. But since $d$ does not appear in $T$ or $\psi$, we have $T\models(\exists v\,\varphi(v))\rightarrow \psi$, as was to be shown.