Let $X, Y$ be topological spaces and $Op(X), Op(Y)$ be corresponding categories of open sets, where morphisms are given by inclusions. For any continuous and open function $f:X\to Y$, there exists a functor $F_{f}:Op(Y) \to Op(X)$ defined as $F_{f}(U) := f^{-1}(U)$, which is continuous in the sense that it preserves limits: $$ F_{f}\left(\lim_{i} U_{i}\right) \simeq \lim_{i} F_{f}(U_{i}). $$ (This equation is equivalent to $$ f^{-1} \left( \left( \bigcap_{i} U_{i}\right)^{\circ}\right) = \left(\bigcap_{i} f^{-1}(U_{i})\right)^{\circ} $$ which is true if $f$ is both continuous and open. If $f$ is not open, taking iterior may not commutes with takeing inverse image.)
I want to know if there's a sort of a converse of this theorem: for given $G:Op(Y) \to Op(X)$, there exists unique $g:X\to Y$ which is continuous (and maybe also open) and $G \simeq F_{g}$. If $X =Y= \{a, b\}$ are topological spaces with an indiscrete topology, then $f_{1} = \mathrm{id}$ and $f_{2}(a) =b, f_{2}(b)=a$ induces same $F_{f_{1}} = F_{f_{2}}$. I want to reconstruct $f$ from $G:Op(Y) \to Op(X)$ by imposing some condition on $G, X,$ and $Y$. It seems like the theorem is true if $X, Y$ are both Hausdorff, but I'm not sure about this.
This kind of stuff is usually called pointless topology (or point-free topology, or locale theory). There is a lot to say about this, and it is probably best to refer you to Sheaves in Geometry and Logic by Mac Lane and Moerdijk to read about it. In particular, chapter IX is all about locales. Of course, it goes into topos theory as well, but you can skip that bit if you wish.
Here are a few results that might interest you (the numbering refers to the book I just mentioned).
So given any continuous function of topological spaces $f: X \to Y$, we have a corresponding lattice homomorphism $f^{-1}: Op(Y) \to Op(X)$, which gives us a morphism of locales in the other direction: $f^{-1}: Op(X) \to Op(Y)$. An actual point $x \in X$ of a topological space is just a continuous function $p_x: * \to X$ where $*$ is the one-point space. The locale corresponding to $*$ is $\{0,1\}$, so this translates into a morphism of locales $p_x: \{0,1\} \to Op(X)$.
So one might hope that the set $pt(L)$ of points on a locale (i.e. all the morphisms $\{0,1\} \to L$) can be made into a topological space such that $Op(pt(L)) \cong L$. As you already noted, this is generally not true. However, this is part of an adjunction.
Of course, I have not told you here how to put a topology on $pt(L)$, but for the details I refer to the book I mentioned (the construction is right before the cited theorem).
You already noted that requiring our spaces to be Hausdorff gives hope of an actual equivalence of categories. In fact, we need a little less, namely just that the spaces are sober. Any Hausdorff space is sober.
("locales with enough points", another definition I refer you to the book for)
This is getting very close to what you are asking about, but you wanted to know about open continuous functions. For this we have to look a bit further into the chapter, where we find section IX.7 "Open Maps of Locales". An important thing to note is that an open continuous function $f: X \to Y$ gives us a pair of adjoints $f_!: Op(X) \leftrightarrows Op(Y): f^{-1}$ (and then it immediately follows that $f^{-1}$ preserves limits, as you mentioned). Another nice property they always satisfy is the Frobenius identity: $$ f_!(V \cap f^{-1}(U)) = f_!(V) \cap U, $$ where $V \subseteq X$ and $U \subseteq Y$ opens. Taking these properties as a definitions for an open map of locales (def IX.7.1) we have the following proposition.
Note that any Hausdorff space is $T_1$. So if we restrict our interest to Hausdorff spaces, then we are indeed able to recover a space from its locale of opens and the open continuous functions can be characterised just from induced maps of locales (i.e. we can say whether or not $f$ is open based on $f^{-1}$).