This is a follow-up on this post, which is based upon this paper. First, let me set up some definitions, etc.
A Semigroup $S$ is said to be an inverse semigroup provided that for every $x \in X$, there exists a unique element $x^{-1}$ such that $x^{-1}xx^{-1} = x^{-1}$ and $xx^{-1}x=x$.
The semilattice $E$ of idempotents in an inverse semigroup $S$ is given by $E = \{x^{-1}x \mid x \in S\} = \{xx^{-1} \mid x \in S\} = \{e \in S \mid e = e^2 \}$.
Given a set $X$, the collection of all partial bijections on $X$ is denoted as $I(X)$. In the MSE post I linked above, we determined that the semilattice of idempotents in $I(X)$ consists of those partial bijections $f$ such that there exists a subset $A \subseteq X$ and $f : A \to A$ is the identity. This means that partial bijections in the semilattice can be identified with subsets of $X$.
Now, let's consider the case when $X = P$ is a left-cancellative monoid. Let $I_{l}(P) \subseteq I(P)$ denote the inverse semigroup generated by the partial bijections $P \to pP$, $x \mapsto px$. On page 18 of the paper I linked above, the author defines $\mathcal{J}_{P}$ as the semilattice of idempotents in $I_{l}(P)$, and claims that it is easy to see that
$$\mathcal{J}_{P} = \{p_n ... q_1^{-1} p_1(P) \mid q_i,p_i \in P\} \cup \{q_n^{-1} p_n ... q_1^{-1} p_1 (P) \mid q_i, p_i \in P \},$$
where, for $X \subseteq P$ and $p \in P$, $p(X) := \{px \mid x \in X\}$ and $p^{-1}(X) := \{y \in P \mid qy \in X\}$. I, however, don't see why $\mathcal{J}_{P}$ consists of all such sets.