$\DeclareMathOperator\cp{cp}$We will derive some additional necessary conditions from the following
Observation: Let $\tau$ be a topology on $X$ and $\tau'$ a topology refining $\tau$. Suppose that $(X,\tau')$ is compact. Then any $\tau'$-closed set is $\tau$-compact.
Indeed, it is compact in $\tau'$ because it is closed in a compact, and so it is compact also in $\tau$ because the identity $\tau' \to \tau$ is continuous.
Consequences: Let $(X,\tau)$ be a topological space admitting a $\beta$-structure $\tau^\xi$. Then:
$(X,\tau)$ is compact (as noted in the question).
$(X,\tau)$ is locally compact (in the sense that for every $x \in X$ there is a local base of compact neighborhoods). This follows from condition (3) on a $\beta$-space and the Observation.
$(X,\tau)$ is "c-separated": For every disjoint $C,D \subseteq X$ which are either closed or singletons, there exist compact $K,L \subseteq X$ such that $C \cap K = \emptyset$, $D \cap L = \emptyset$, and $K \cup L = X$. This follows from the fact that $(X,\tau^\xi)$ is Hausdorff, regular, and normal and the Observation.
$(X,\tau)$ is "c-completely separated": Let $C,D \subseteq X$ be disjoint and either closed or singletons. Then there exists a (not necessarily continuous) function $f: X \to [0,1]$ such that $f^{-1}(0) = C$, $f^{-1}(1) = D$, and $f^{-1}([a,b])$ is compact for every $a \leq b$. This follows from the fact that $(X,\tau^\xi)$ has the corresponding separation property and the Observation.
Note also that if the collection of sets with compact complement forms a topology, this this topology is the unique $\beta$-structure on $(X,\tau)$. But this is not necessarily the case.
