name for monoids inducing bimonoids in Rel? Let Rel be the category of sets and relations, which is a (compact closed) symmetric monoidal category under the cartesian product of sets. We write $A \nrightarrow B$ to indicate a relation from $A$ to $B$, and given a function $f : A \to B$, we write $f^+ : A \nrightarrow B$ and $f^- : B \nrightarrow A$ for the corresponding functional/anti-functional relations.
Now, any ordinary monoid $(X,m,e)$ (i.e., monoid in Set, where $m : X \times X \to X$ is the multiplication and $e : 1 \to X$ is the unit) induces a pair of a "relational monoid" $(X,m^+,e^+)$ and a "relational comonoid" $(X,m^-,e^-)$, that is, an object $X$ which is both a monoid and a comonoid in Rel. I am interested in when $(X,m^+,e^+,m^-,e^-)$ forms a bimonoid in Rel.
The condition that $(X,m^+,e^+,m^-,e^-)$ is a relational bimonoid reduces to the following conditions on the original set-theoretic monoid (writing $m(x,y)$ as $x \cdot y$):


*

*$x\cdot y = x'\cdot y'$ if and only if there exist $t,u,v,w$ such that $x = t\cdot u$ and $y = v\cdot w$ and $x' = t\cdot v$ and $y' = u\cdot w$.

*If $x\cdot y = e$ then $x = e$ and $y = e$.


Question: Do monoids satisfying conditions (1) and (2) have a name?
 A: I think describing all monoids satisfying these conditions is an essentially impossible task. But for a finite band (monoid where each element is idempotent) this condition is equivalent to being a distributive lattice with meet as the binary operation. 
As observed by Zeilberger in the comments $M$ must be commutative. If all its elements are idempotent, then it is a meet semilattice with top via the order $e\leq f$ iff $ef=e$. The product is the meet. By finiteness any set with a lower bound has a meet. Everything I'm about to write is true for meet semilattice with this property (ie completeness). Actually all I need is a top and binary joins. 
It was observed in the comments if $M$ is a distributive lattice with top, then it satisfies the above properties (the commenter used the dual convention of join as the operation). 
If $M$ is a meet semilattice which is either finite or complete then since it has a top it has joins. So it is a lattice. From now on just assume $M$ is a lattice with top. If it is not distributive it contains a sublattice isomorphic to $M_3$ or $N_5$. See https://en.m.wikipedia.org/wiki/Distributive_lattice
If it contains $M_3$ then we can find distinct $x,y,x'$ in $M$ so that any two of these elements have the same join $m$ and same meet. From $xy=x'y$ we get $x=tu$, $x'=tv$ and $y=uw=vw$. Thus $t,u,v$ are each greater than the join $m$. This contradicts the given equalities.
Similarly, if $M$ has a sublattice isomorphic to $N_5$ we can find $x,x',y$ with $x'<x$ and $xy=x'y$ such that the join $m$ of $x',y$ is the join of $x,y$. From the condition we then have  $x=tu$, $x'=tv$ and $y=uw=vw$ and so $u,v\geq m\geq x$. This contradicts $x'=tv$ as $t\geq x$. 
