Kong Liang told me a proof for
the statement **if both kernels and images of $d_n$ are Abelian groups then all the monoids $M_n$ are Abelian groups**.

Let  $f: A \to B$ be a surjective morphism between two commutative monoids.
Surjectivity means  $B =$ Img$(f)$.
Assume that Ker$(f)$ is an Abelian group and $B =$ Img$(f)$ is an Abelian group.

Then $A$ is automatically an Abelian group.

Proof:

1. it is enough to show that any element $a$ in $A$ has a right inverse $a'$,
i.e.  $a a' = 1$. (by the commutativity,  $a' a = a a' = 1$.)
Notice that if such $a'$ exists, it must be unique. Otherwise,
let $a'$ and $a''$ be such that     $a a' =1 = a a''$.
Then we have $a' = a' a a'' = a''$  (use commutativity)

2. For any $a$ in $A$, let $b$ be the inverse of $f( a )$
and let $c$ be an element in $A$ such that $f( c ) = b$. Then we have
$f(a c) = f( a ) f( c ) = f( a ) b =1$.
Therefore, $a c$ is in the Ker$(f)$. Since Ker$(f)$ is an Abelian group,
there is an element $d \in A$ such that $acd = 1$.
Hence $cd$ is the right inverse of $a$. 

Therefore the quasi-cochain complex is the usual cochain complex, and there is a lot of work on cochain complex, as well as a lot of theorems. :-)
Thanks, @ Fernando Muro !