TITLE: flat maps of monoids which are not localizations
QUESTION [5 upvotes]: It is well known that a localization $S^{-1}R$ of a commutative ring $R$ is flat as a $R$-module.
Rather, I am looking for extensions of rings which share certain properties of localizations, like flatness, while inverting nothing.
More precisely, I would like to find classes of examples, if such exist, of maps of monoids $f:M\to N$ verifying

$N^{\times} = \{1\}$,
$f$ is injective, but not surjective,
"$M$ is cofinal in $N$" in the sense that the augmentation induces an isomorphism $\Bbb{Z}[N]\otimes_{\Bbb{Z}[M]}\Bbb{Z}[1]\cong\Bbb{Z}[1]$ of abelian groups (note that $\Bbb{Z}[1]\cong\Bbb{Z}$ as abelian groups),

for which $\Bbb{Z}[N]$ is flat as either a left or right $\Bbb{Z}[M]$-module.
The first two conditions listed above are simply to guarantee that $f$ does not invert any elements whatsoever, while not being an isomorphism.
For completeness, and in case it is easier, I am actually looking for examples of maps $f:M\to N$ for which $\text{Tor}^{\Bbb{Z}[M]}_\ast(\Bbb{Z}[N],\Bbb{Z}[N])$ is $\Bbb{Z}[N]$ concentrated in degree zero.
Remark: this condition is related to the forgetful functor from the positive derived category of $\Bbb{Z}[N]$-modules to the derived category of $\Bbb{Z}[M]$-modules being full and faithful.
Edit: I have removed the extraneous commutativity condition on the monoids.

REPLY [2 votes]: This is not really an answer, but it is too long for a comment. Suppose that $f\colon M\to N$ is a homomorphism of monoids (not necessarily commutative).  Let's work with right modules.
First of all condition (3) can be phrased as follows: construct a graph with vertex set $N$ and an edge between $n$ and $nm$ for all $n\in N$ and $m\in M$.  Then (3) happens iff this graph is connected.  
Next there is a classical condition coming from category/topos theory that guarantees flatness of $\mathbb ZN$ over $\mathbb ZM$.  In fact, it is equivalent to flatness of $N$ as a right $M$-set, but this is a stronger condition.  Namely, consider the category of elements $\mathcal C$ of $N$ viewed as a presheaf over $M$.  So the objects are elements of $N$ and there is an arrow $(n,m)\colon nm\to n$ associated to each $n\in N$ and $m\in M$.  Composition is $(n,m)(nm,m')=(n,mm')$. 


Fact. If the category $\mathcal C$ is filtered one has that $\mathbb ZN$ is flat over $\mathbb ZM$.