The monoid of fractions associated with the submonoid of cancellable elements of a commutative monoid E Let $E$ be a commutative monoid, $\Sigma$ the submonoid of cancellable elements of $E$, $E_{\Sigma}$ the monoid of fractions of $E$ associated with $\Sigma$ and $\varepsilon$ the canonical homomorphism of $E$ into $E_\Sigma$. 
Since every element of $\Sigma$ is cancellable, $\varepsilon$ is injective.
Logically speaking, what does "identifying the monoid $E$ with the submonoid $\varepsilon(E)$ of $E_\Sigma$" allow me to do? I mean: w.r.t the set-theoretic formulas, what can I formally replace with what? For example, does it allow me to write $\varepsilon(a)=a$ for $a\in E$ (I hope not...) ? What is gained by this "identification"? 
I ask this because I would like to know how one can "identify" $E_{\Sigma}$ with the submonoid generated by $E\cup\Sigma^*$, where $\Sigma^*$ denotes the set of inverses of the elements of $\Sigma$.
EDIT: (You can ignore the details concerning this specific construction of the monoid of fractions)

 A: General remarks
If you have algebraic structures $A$ and $B$, and an injective map $f\colon A\to B$ that is a morphism, then when we say that we "identify $A$ with a substructure of $B$" we mean that we can consider the function $f$ as an isomorphism onto its image, and then consider the inclusion map. That is, $A\stackrel{f}{\to}B$ factors as 
$$A\stackrel{f,\cong}{\longrightarrow } f(A)\stackrel{i}{\hookrightarrow} B.$$
We commonly "identify" two structures when they are isomorphic: the isomorphism just represents a "renaming" of elements. You identify $a\in A$ with $f(a)\in f(A)$; the fact that $f$ is one-to-one ensures that this is simply a relabeling of names. The fact that $f$ is a homomorphism guarantees that the algebraic structure is being preserved.
Once you accept that $f(A)$ is "essentially" just $A$, then you can follow it up by the embedding, thus we have recognized ("identified") $a$ with its image inside of $B$. Given that there is a bijection between $A$ and $f(A)$ that respects the structure, one may just as well "skip the intermediary" and simply work with $f(A)$ sitting inside of $B$, rather than with the three distinct objects $A$, $f(A)$, and $B$. 
Thus, for example, even though $\mathbb{N}$ is not technically a subset of $\mathbb{Z}$ (as the latter is constructed as a set of equivalence classes of pairs of natural numbers), there is a natural embedding of $\mathbb{N}$ into $\mathbb{Z}$ and we treat the image as if it were $\mathbb{N}$ itself, thus "identifying" $\mathbb{N}$ with its canonical image inside of $\mathbb{Z}$ and treating $\mathbb{N}$ as just a subset/substructre of $\mathbb{Z}$. 

Here you start with a monoid $E$. You construct a separate, distinct monoid $E_{\Sigma}$. But you would really like to think of $E_{\Sigma}$ as being an "extension" of $E$, something you get out of $E$ by "adding" stuff to it, just like we think of $\mathbb{Q}$ as being an extension of $\mathbb{Z}$, even though the latter are not "fractions". 
So you have a morphism $\varepsilon E\to E_{\Sigma}$ that is one-to-one. That allows you, as per above, to think of $E$ as being a subset of $E_{\Sigma}$ via the embedding, and so ignore the technical fact that $E$ is not literally a submonoid of $E_{\Sigma}$ (it's not even a subset). 
What is gained is purely conceptual: you can now think of $E_{\Sigma}$ as being an extension of $E$ obtained by adding stuff to it. Just like thinking of $\mathbb{Z}$ as a subset of $\mathbb{Q}$: you can formally do everything via a one-to-one morphism, but it is clearer and less cluttered to simply think of $\mathbb{Z}$ as a subset of $\mathbb{Q}$, rather than to think of the map that sends $\mathbb{Z}$ into $\mathbb{Q}$ and respects the operations. 
The final paragraph actually shows you why this identification is useful: you have $S\subseteq \Sigma$. You can construct two different monoids, $E_S$ and $E_{\Sigma}$. Formally, they are completely separate objects, with different underlying sets, and with a different equivalence relation defined on them. But you want to establish a relation between $E$, $E_S$, and $E_{\Sigma}$. 
By thinkin of $E$ not as a separate object distinct from $E_{\Sigma}$, but rather as a subobject of $E_{\Sigma}$, you can also view $S$ as a subset of $E_{\Sigma}$; now you have the set $E\cup S$ sitting inside $E_{\Sigma}$, and so you can consider the submonoid it generates. Then one obtains a morphism from $E_S$ to this submonoid and proves that this is one-to-one, so that you can actually think of these three monoids as "sitting inside one another",
$$E \subseteq E_S\subseteq E_{\Sigma}$$
instead of thinking of them as three completely separate objects, $E$, $(E\times S)/R_S$, and $(E\times \Sigma)/R_{\Sigma}$ (where $R_S$ is the corresponding equivalence relation for the construction of $E_S$ and $R_{\Sigma}$ for the construction of $E_{\Sigma}$.
Just like you can construct a ring starting from $\mathbb{Z}$ and formally adding a multiplicative inverse of $2$; and then consider the rationals as being obtained by inverting all nonzero elements. Formally, three different rings ($\mathbb{Z}$, $\mathbb{Z}[\frac{1}{2}]$, and $\mathbb{Q})$, but you'd rather think of them as sitting insider one another, $\mathbb{Z}\subseteq \mathbb{Z}[\frac{1}{2}]\subseteq \mathbb{Q}$. Otherwise, you need to keep track of all the embedding morphisms and you will make talking about these objects as they related to one another extremely cumbersome. 
A: Whether or not $\epsilon(a) = a$ depends on the set theoretic construction of the monoid of fractions, but in practice the answer is either no or it's irrelevant.
More formally you can say the monoid $E$ is isomorphic to the monoid $\epsilon(E)$. Informally it means that the monoid structures of $E$ and $\epsilon(E)$ are the same, so properties of the monoid structure that don't depend on the precise set-theoretic construction are the same in both $E$ and $\epsilon(E)$. This is not a precise answer. I believe there is a precise answer, but it would be a very long answer.
I don't understand the second part of the question. $E \cup\Sigma^{*} = E$, since $\Sigma^{*} \subseteq E$
