Is there a name for a regular semigroup with zero in which the product of any two different idempotents is zero?

For what it's worth here is a characterisation of the finite semigroups that are regular with zero in which the product of any two different idempotents equals zero.

If $S$ and $T$ are semigroups with zero, then the 0-union of $S$ and $T$ is the semigroup formed by taking the union of $S$ and $T$ but identifying their zero elements, and with multiplication extending that of $S$ and $T$ by $st=ts=0$ for all $s\in S$ and $t\in T$.

A finite Brandt semigroup $S$ is a finite 0-simple inverse semigroup (a semigroup where there are no proper non-zero ideals, and where every element $x\in S$ has a unique $x^{-1}\in S$ such that $xx^{-1}x=x$ and $x^{-1}xx^{-1}=x^{-1}$).

A finite regular semigroup $S$ with zero has the property that the product of any two distinct idempotents equals zero if and only if it is a 0-union of Brandt semigroups.

For the converse, let $B_i$, $i\in \{1,\ldots, n\}$ for some $n\in \mathbb{N}$ be Brandt semigroups, and let $S$ denote the 0-union of the $B_i$. Then $S$ is an inverse semigroup, and hence regular. The $\mathscr{J}$-classes of $S$ are simply $\{0\}$, and $B_i\setminus\{0\}$ for all $i$. Suppose that $e\in B_i$ and $f\in B_j$ are idempotents, and $e\not=f$. If $i\not=j$, then $ef=0$ by definition of the multiplication in the 0-union. If $i=j$, then $ef$ is the product of distinct idempotents in a $\mathscr{J}$-class of a finite inverse semigroup, and so $ef\not\in B_i$. Thus $ef=0$, as required.

Let $S$ be a finite regular semigroup with the property that the product of any two distinct idempotents equals $0$. We start by showing that $S$ is an inverse semigroup. It suffices to show that every $\mathscr{L}$- and $\mathscr{R}$-class of $S$ contains exactly one idempotent. Since $S$ is regular, every $\mathscr{L}$- and $\mathscr{R}$-class contains at least one idempotent.

Suppose that $e\in S$ is a non-zero idempotent. If there exists $f\in S$ such that $e\mathscr{L}f$, then there exists $x\in S$ such that $xe=f$. Hence $x\cdot ef=f^2=f$ and so $ef\mathscr{L}f$, and, in particular, $ef\not=0$. This contradicts the assumption, and hence every $\mathscr{L}$-class of $S$ contains exactly one idempotent. A dual argument shows that every $\mathscr{R}$-class of $S$ contains exactly one idempotent, and so $S$ is an inverse semigroup.

Suppose there are non-zero elements $x,y\in S$ such that $S^1xS^1\subsetneq S^1yS^1$. Let $J_x$ and $J_y$ denote the $\mathscr{J}$-classes of $x$ and $y$ in $S$, respectively. Since $S$ is regular, there exist idempotents $e\in J_y$ and $f\in J_x$, and there exist $u,v\in S$ such that $f=uev$. Hence $fv^{-1}\mathscr{R}f$ and $u^{-1}f\mathscr{L}f$, and so, since $\mathscr{L}$ is a right congruence, $u^{-1}fv^{-1}\mathscr{L}fv^{-1}\mathscr{R}f$, i.e. $u^{-1}fv^{-1}\mathscr{D}f$. In particular, $u^{-1}fv^{-1}\not=0$. But $u^{-1}fv^{-1}=(u^{-1}u)e(vv^{-1})$ is a product of idempotents, which does not equal $0$. It follows that $u^{-1}u=vv^{-1}=e$, which implies that $u^{-1}fv^{-1}=e$ and so $e\mathscr{J}f$, a contradiction.

It follows that the union of any $\mathscr{J}$-class and $\{0\}$ is a subsemigroup of $S$, which is $0$-simple and inverse, i.e. a Brandt semigroup. From the previous paragraph, it follows that $S$ is the 0-union of these Brandt semigroups, which completes the proof.