Is there a (discrete) monoid M injecting into its group completion G for which BM is not homotopy equivalent to BG?

For a (discrete) monoid $M$, the classifying space $BM$ is the
geometric realization of the nerve of the one object category whose
hom-set is $M$. (This definition gives the usual classfiying space
when $M$ is a group.) The group completion of $M$ can be constructed
as the fundamental group of $BM$, and is characterized by the
universal property that any monoid homomorphism from $M$ to a group
factors uniquely through the group completion.

My question is whether there is an example of a monoid for which the
canonical map to its group completion is injective, but for which this
canonical map does not induce a homotopy equivalence of the
classifying spaces.

As background here are some facts:

1. Classifying spaces of monoids produce all connected homotopy types!
   This is proved in Dusa McDuff’s 1979 paper On the classifying spaces
   of discrete monoids. For a neat concrete example, see Zbigniew
   Fiedorowicz’s A counterexample to a group completion conjecture of
   JC Moore; it shows a specific 5
   element monoid whose classifying space is homotopy equivalent to $S^2$.

2. If $G$ is the group completion of a commutative monoid $M$, the
   canonical map $BM to BG$ is a homotopy equivalence; even if $M to G$
   is not injective. (This is easy to prove: think of $M to G$ as a
   functor between one object categories and apply Quillen’s Theorem A to
   it. There is only one slice category to check and using commutativity
   it is easy to see this category is filtered and thus contractible.)

3. If $M$ is a free monoid and the free group $G$ is its completion,
   the map $BM to BG$ is a homotopy equivalence. It fact, more
   generally, if $C$ is the free category on some directed graph $X$, the
   nerve of $C$ is homotopy equivalent to the geometric realization of
   $X$. This is proved in Dwyer and Kan’s Simplical Localization of
   Categories,
   proposition 2.9, but the proof is simple enough to sketch here: for
   each $k$, the inclusion of the $k$-skeletion of $NC$ into the
   $(k+1)$-skeleton is a weak homotopy equivalence (since you get the
   $(k+1)$-skeleton by filling in some horns); so the $1$-skeleton,
   $X$, is weakly equivalent to $NC$. (The claim for free monoids is
   the case where $X$ consists of a single vertex with some loops.)

4. Even if a monoid has left and right cancellation the canonical map
   to its group completion might not be injective. Here’s an example from
   Malcev’s On the Immersion of an Algebraic Ring into a Field: let $M$
   be the monoid presented by $(a,b,c,d,x,y,u,v : ax=by, cx=dy,
   au=bv)$. Malcev shows that $M$ is cancellative, but that in $M$, $cu
   neq dv$; in any group the relations listed for $M$ would imply that
   $cu=dv$.