Consider any finite commutative semigroup $S$.
Say that $x \leq y$ iff $x = y$ or $xy = y$. This is a partial order on $S$: the only nontrivial property to check is that $x \leq y$ and $y \leq z$ implies $x \leq z$, and (WLOG assuming $x \neq y \neq z$) we have $xy = y$ and $yz = z$, so $xz = x(yz) = (xy)z = yz = z$, as desired.
Is the resulting poset always a lattice, after adding a greatest and/or least element if needed?
Here is a counterexample: consider the set $\{0, 1, -1, x, y, \infty\}$. Let $0, 1, -1$ be a subgroup isomorphic to $(\mathbb{Z}/3, +)$, and let $0, x, y, \infty$ be a subsemilattice isomorphic to $(\mathcal{P}(\{x, y\}), \cup)$. We can combine these two commutative semigroups into one big one by letting $x, y, \infty$ absorb $0, 1, -1$. We have that $1 \leq x$, $1 \leq y$, $-1 \leq x$, $-1 \leq y$, but the lower bounds of $x$ and $y$ are precisely $0, 1, -1$, and in particular, there is no greatest lower bound.
