Let $H$ be a (multiplicatively written) commutative monoid with identity $1_H$. Given $X, Y \subseteq H$, we take
$$XY := \{xy: x \in X,\, y \in Y\}.$$
We call a set $I \subseteq H$ an *ideal* of $H$ is $I = IH$.
The set $\mathcal I^\ast(H)$ of all *non-empty* ideals of $H$ is made into a commutative, reduced monoid by endowing it with the binary operation $$\mathcal I^\ast(H) \times \mathcal I^\ast(H) \to \mathcal I^\ast(H): (I, J) \mapsto IJ.$$ Let $\mathcal R$ be the class of all monoids $K$ for which there exists a commutative monoid $H$ such that $K$ is isomorphic to $\mathcal I^\ast(H)$. A couple of days ago, I asked a question ultimately aimed at shedding light on the nature of the class $\mathcal R$, and an answer of Benjamin Steinberg resulted into a number of observations about the objects that sit in $\mathcal R$. But some of these observations were seriously flawed (and they were all mine).

In particular, I stated in the comments to the same answer that (a close relative of) $\mathcal I^\ast(H)$ is unit-cancellative (see the notes at the end of this post for terminology), as I overlooked a trivial case in the argument I pretended to use in the proof: What the argument does actually show is that, if $I, J \in \mathcal I(H)$ and $IJ = I$, then $J = I$ (and, hence, $I$ is an idempotent ideal). So, it seems plausible to me that $\mathcal I^\ast(H)$ *will be* unit-cancellative if $H$ is not "too wild", which leads to the following:

Q.Is it true that $\mathcal I(H)$ is unit-cancellative if so is $H$? If not, what about the (much more restrictive) case when $H$ is cancellative?

*Notes.* A commutative monoid $H$ with identity $1_H$ is *unit-cancellative* if $xy=x$, for some $x, y \in H$, implies $y \in H^\times$, where $H^\times$ is the *set of units* of $H$; *reduced* if $H^\times = \{1_H\}$; *cancellative* if the function $H \to H: x \mapsto ax$ is injective for every $a \in H$ (of course, if $H$ is cancellative, then it's also unit-cancellative).