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:
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)
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 !