Recall that if $M,N$ are two structures of the same type, then $M$ is $\mathcal{L}_{\infty,\omega}$ elementarily equivalent to $N$ precisely when $M$ and $N$ are isomorphic in some forcing extension. I am wondering if there is a topological version of this result.

A frame is a complete lattice $L$ that satisfies the infinite distributivity identity $$x\wedge\bigvee_{i\in I}y_{i}=\bigvee_{i\in I}(x\wedge y_{i}).$$ If $(X,\mathcal{T})$ is a topological space, then $\mathcal{T}$ is always a frame. Frames are the central object of study in point-free topology since all of the information in a good space $(X,\mathcal{T})$ is contained in the lattice $\mathcal{T}$. Most notions from general topology immediately generalize seamlessly the point-free topology.

Suppose that $L$ is a completely regular frame and $B$ is a complete Boolean algebra. Then let $L^{+B}$ be the set of all frame homomorphisms $\phi:L\rightarrow B$. Then set $\|\phi=\theta\|\geq b$ precisely when $\phi(x)\wedge b=\theta(x)\wedge b$ for all $x\in X$. Then $L^{+B}$ is a complete $B$-valued set. $L^{+B}$ becomes a $B$-valued topological space with $B$-valued basis consisting of all systems $\sum_{b\in p}\underline{x_{b}}\cdot b$ where $p$ is a partition of $B$, $x_{b}\in L$ for $b\in p$, and where $\|\phi\in\sum_{b\in p}\underline{x_{b}}\cdot b\|=\bigvee_{b\in p}(\phi(x_{b})\wedge b)\|.$

Suppose $L,M$ are completely regular frames. Then does there exist some logic $\mathcal{L}$ where $L$ and $M$ are $\mathcal{L}$-elementarily equivalent precisely when there is some complete Boolean algebra $B$ such that for all complete Boolean algebras $C$ with $B\subseteq C$ and where $B$ is a complete subalgebra of $C$ we have $L^{+C}=M^{+C}$? Does there exist a version of this result for uniform frames of for complete uniform frames?