Suppose I have a $\land$-semilattice, equipped with a bound $\top$ and an absorbing element $0$ (such that $\forall a.a\land 0 = 0$)
Based on working through a few finite examples, it seems like this structure will always a bounded lattice where $0=\bot$ - albeit without any nice formulaic definitions of $\lor$.
Can this be shown to be true in general?
The result is true if the semilattice $S$ is finite, or, more generally, if it doesn't have infinite descending chains, so that meets of non-empty (finite or infinite) sets always exist.
In that case, we just define, for $a,b \in S$, $$a \vee b = \bigwedge U_{a,b},$$ where $$U_{a,b} = \{ s \in S : a\leq s \,\text{ and }\, b\leq s \}$$ (that is, $U_{a,b}$ is the set of upper bounds common to $a$ and $b$).
To see that this doesn't happen in the general case, consider the diagram below.
(The elements above the dots are supposed to be above both $a$ and $b$)
Here we have a bounded $\wedge$-semilattice which is not a lattice since there doesn't exist the element $a\vee b$.
Notice this is because the set $U_{a,b}$ has no minimum element.