Construction of the monoid of fractions This question comes from my attempt to solve Exercise 17(b) of Bourbaki, Algebra, Chapter 1, §2. 
Let $E$ be a commutative monoid (written multiplicatively) and $S$ a submonoid of $E$. Define on $E\times S$ an equivalence relation by $(a,s)\sim(b,t):\Leftrightarrow$ "$\exists u\in S$ such that $atu=bsu$". Denote the set $(E\times S)/\sim$ by $\overline{E}$ and the class of $(a,s)$ by $a/s$. For any $a\in E$, let $\epsilon(a)=a/e$.
So far so good. The Bourbaki exercise is actually more general (replacing commutativity by weaker assumptions), but, even in this special case, I can't do the next step:
Show that there exists on $\overline{E}$ one and only one monoid structure such that $\epsilon$ is a monoid homomorphism and such that, for all $s\in S$, $\epsilon(s)$ is invertible. 
My problem is with the "only one" part. Given a monoid structure $\otimes$ on $\overline{E}$ with those properties, I see no reason why $(a/s)\otimes(b/t)=(ab)/(st)$ or even $(s/e)\otimes(e/s)=e/e$. I couldn't find a counterexample in the case $E=\mathbb{N}$, nor did the search give me any intuition on why the assertion should be true.
This commutative case is treated in the text of Bourbaki's Algebra, but there no mention is made of "only one".
I'm glad for anything that gets me started.  
 A: I think that the "only one" part does not hold. Usually there is plenty of monoid structures on the set $\bar{E}$ such that $\epsilon$ is a monoid homomorphism and $\epsilon(s),s\in S$ are invertible.
For example, let $E=S=\mathbb{N}$ be the natural numbers with its usual addition. Then $\bar{E}=\mathbb{Z}$ is essentially the integers with its usual addition which is denoted by $+$. Let $\phi$ be any non-identity permutation of the integers which fixes the nonnegative integers. Then $m\oplus n=\phi^{-1}(\phi(m)+\phi(n))$ defines a new monoid structure $\oplus$ on the (set of) integers, which has the two required properties.
First let us check that $\oplus$ is an associative and commutative operation which has $0$ as its neutral element; if $m,n,p \in \mathbb{Z}$, then
$$
\begin{eqnarray*}
(m\oplus n)\oplus p &=& \phi^{-1}(\phi(m)+\phi(n))\oplus p \\
&=& \phi^{-1}(\phi\phi^{-1}(\phi(m)+\phi(n))+\phi(p)) \\
&=& \phi^{-1}((\phi(m)+\phi(n))+\phi(p)) \\
&=& \phi^{-1}(\phi(m)+(\phi(n)+\phi(p)) \\
&=& \phi^{-1}(\phi(m)+\phi\phi^{-1}(\phi(n)+\phi(p))) \\
&=& m \oplus \phi^{-1}(\phi(n)+\phi(p)) \\
&=& m\oplus (n\oplus p),
\end{eqnarray*}
$$
$$
m\oplus n=\phi^{-1}(\phi(m)+\phi(n))=\phi^{-1}(\phi(n)+\phi(m))=n\oplus m
$$
and
$$
m\oplus 0 = \phi^{-1}(\phi(m)+\phi(0)) = \phi^{-1}(\phi(m)+0)=\phi^{-1}(\phi(m))=m.
$$
Then let us check that $\epsilon:(\mathbb{N},+)\to(\mathbb{Z},\oplus),\epsilon(n)=n$ is a monoid homomorphism; if $m,n\in \mathbb{N}$, then
$$
\epsilon(m) \oplus \epsilon(n) = m \oplus n = \phi^{-1}(\phi(m)+\phi(n)) = \phi^{-1}(m+n)=m+n=\epsilon(m+n);\ \epsilon(0)=0.
$$
Each of the elements $\epsilon(n),n\in\mathbb{N}$ is invertible since for each $n\in\mathbb{N}$
$$
n\oplus \phi^{-1}(-\phi(n)) = \phi^{-1}(\phi(n)+\phi\phi^{-1}(-\phi(n)))=\phi^{-1}(\phi(n)+(-\phi(n))=\phi^{-1}(0)=0.
$$
Lastly note that $\oplus$ is different from $+$ because there exists such a positive integer $m$ that $\phi(-m)\neq -m$, which implies that $m+\phi(-m)\neq 0$ and
$$
m\oplus(-m)=\phi^{-1}(\phi(m)+\phi(-m))=\phi^{-1}(m+\phi(-m))\neq 0\quad (=m+(-m)).
$$

Thank you to Theo Buehler for suggesting a correction to my initial flawed attempt to answer the question.
