A monoid $M$ satisfies the left Ore condition if for all $a,b \in M$ there exist $c,d \in M$ such that $ca=db$. Suppose, in addition, $ac=bc$ implies that there exists an element $d \in M$ such that $da=db$. Consider the following equivalence relation $ \sim$ on $M \times M$, for all $a,b,c,d \in M$, \begin{align} (a,b) \sim (c,d)\hspace{0.1cm} \Leftrightarrow \exists x,y \in M \hspace{0.1cm}\text{such that}\hspace{0.1cm} xa=yc, \hspace{0.1cm}xb=yd. \end{align}
The equivalence class of $(a,b) \in M \times M$ is denoted by $a^{-1}b$. The quotient set $$M^{-1}M=\lbrace a^{-1}b\mid a,b \in M\rbrace$$ is called the group of left fractions of $M$. This is a group for the operation \begin{align} (a^{-1}b)(c^{-1}d)=(xa)^{-1}(yd), \end{align} where $x,y \in M$ are such that $xb=yc$, defines a group structure on $M^{-1}M$.
Similarly, $M$ satisfies the right Ore condition if for all $a,b \in M$ there exist $c,d \in M$ such that $ac=bd$. Suppose, in addition, that $ca=cb$ implies that there exists an element $d \in M$ such that $ad=bd$. Consider the equivalence relation on $M \times M$, where for all $a,b,c,d \in M$ \begin{align*} (a,b) \sim (c,d)\hspace{0.1cm} \Leftrightarrow \exists x,y \in M \hspace{0.1cm}\text{such that}\hspace{0.1cm} ax=cy, \hspace{0.1cm}bx=dy. \end{align*} One then obtains a group $$MM^{-1}=\lbrace ab^{-1}\mid a,b \in M \rbrace,$$ for the operation \begin{align*} (ab^{-1})(cd^{-1})=(ax)(dy)^{-1}, \end{align*} where $x,y \in M$ are such that $bx=cy$.
Suppose all the conditions above hold, such that one can construct both the group of left and right fractions of $M$. I'm trying to construct a straightforward group isomomorphism between $M^{-1}M$ and $MM^{-1}$, but I have trouble finding a map which preserves the group structure. Any help?