I do not believe that there is a nice way to characterize $\Xi$. Furthermore, I do not see any reason to separate sets with least upper bounds or greatest lower bounds from the underlying partial ordering. In this answer, I will give cases of where the partial ordering $\leq$ is completely determined by $\mathcal{A}$ and $\mathcal{B}$. In particular, since $\mathcal{A}$ and $\mathcal{B}$ sometimes completely determine the partial order $X$, we do not obtain a simpler structure when we observe the sets $\mathcal{A}$ and $\mathcal{B}$ as opposed to the partial ordering $\leq$.

Suppose that $X$ is a poset. Let $\preceq^{\circ}$ be the ordering on $X$ where $x\preceq^{\circ}y$ if and only if $A\cup\{x\}\in\mathcal{A}$ implies $A\cup\{y\}\in\mathcal{A}$. Notice how $\preceq^{\circ}$ only depends on $\mathcal{A}$ and not the partial ordering. It is clear that $\preceq^{\circ}$ is a preordering on $X$. Furthermore, if $X$ is a join-semilattice, then $x\preceq^{\circ}y$ whenever $x\leq y$.
Define $\preceq$ to be the relation where $x\preceq y$ if and only if $x\preceq^{\circ}y$ and $x\preceq_{\circ}y$. If $X$ is a lattice, then $x\preceq y$ whenever $x\leq y$. In fact, as the following proposition shows, there are some lattices such that $x\preceq y$ if and only if $x\leq y$.

$\mathbf{Proposition}$. Suppose $X$ is a Heyting algebra such that if $x<y$, then there is some non-empty set $R\subseteq\{z\in X|x\leq z\leq y\}$ that does not have a least upper bound. Then $x\leq y$ if and only if $x\preceq^{\circ}y$.

$\mathbf{Proof}$. We only need to show that if $x\not\leq y$, then $\neg(x\preceq^{\circ}y)$. To prove this result, we will need a simple fact about Heyting algebras. In a Heyting algebra, if $\bigvee R$ exists and $a\in X$, then
$\bigvee_{r\in R}(a\wedge r)$ exists and $a\wedge\bigvee R=\bigvee_{r\in R}(a\wedge r)$. Suppose that $x\not\leq y$. Then $y\wedge x<x$, so there is some non-empty set $R\subseteq\{z\in X|y\wedge x\leq z\leq x\}$ with no least upper bound. Then $\bigvee(R\cup\{x\})=x$, so $R\cup\{x\}\in\mathcal{A}$. However, $R\cup\{y\}\not\in\mathcal{A}$. Otherwise, if $R\cup\{y\}\in\mathcal{A}$, then $\bigvee_{a\in R\cup\{y\}}(a\wedge x)$ exists. However, it is clear that if $\bigvee_{a\in R\cup\{y\}}(a\wedge x)$ exists, then
$\bigvee_{a\in R}(a\wedge x)=\bigvee R$ exists, a contradiction. Therefore we conclude that $\neg(x\preceq^{\circ}y)$.