Suppose that $\left(E, \cdot\right)$ is a commutative semigroup of idempotents then the relation $\leq$ on $E$ defined by $$a\leq b \iff ab =b,$$ is a partial order on $E$. Furthermore, $\left(E,\leq\right)$ is an upper semilattice with the join of $a$ and $b$, $a\vee b$ being their product $ab$ and $ab$ is the unique least upper bound of $a$ and $b$.
I have attempted a proof but I am worried that the claim is false. It is an analogue of Proposition 1.3.2 in J.M. Howie's Fundamentals of Semigroup Theory.
The claim is true. Let us see the relation you defined is a partial ordering.
Let $a,b,c \in E$.
Since $a\cdot a = a$, we have $a \leq a$, so the relation is reflexive.
If $a \leq b$ and $b\leq a$ then $a = b\cdot a = a \cdot b = b$, so it's anti-symmetric.
If $a \leq b$ and $b\leq c$, then $a \cdot c = a \cdot(b \cdot c) = (a \cdot b) \cdot c = b \cdot c = c$, so it's transitive.
If $a\leq c$ and $b \leq c$, then $c = a \cdot c = a \cdot (b\cdot c) = (a\cdot b)\cdot c$, whence $a\cdot b \leq c$, and so $a\cdot b$ is the least element above $a$ and $b$.
Another way of seeing this is that a semi-lattice is precisely an idempotent commutative semi-group, which exactly what you have.