Thinking of the group $(\mathbb{Z}, +)$ as a single-object category? I am totally new to the category theory and my question might sound stupid. Nevertheless, I am eager to understand.
I assume existence of category $C$ with a single object $A$. So each an every arrow in this category deliberately points back to itself. In other words, all the maps there are $A \mapsto A$. 
Now, at this point, nothing guarantees that those $A \mapsto A$ morphism are indeed isomorphisms. It appears to happen only when any group emerges, since, by definition, whatever group required to ship pairs of $(a, a^{-1})$; hence, it is perfectly fine to acknowledge the existence of $a \mapsto a^{-1}$ morphism together with $a^{-1} \mapsto a$, which both make $(a, a^{-1})$ isomorphic to each other. 
Next step is to define, let say, $(\mathbb{Z}, +)$ group as a category $C$. At this point I fail:

Then category $C$ consists of set (or class) $A \mapsto A$ maps and associative composition function $(A, A) \times (A, A) \mapsto (A, A)$.

I can't understand: $(+)$ is not an instance of $(A, A) \times (A, A) \mapsto (A, A)$. So somehow $(+) : (g \in A, g' \in A) \mapsto g'' \in A$ must be "upgraded" to become an $((+, +), (+, +)) \mapsto (+, +)$ and I see no clear yet obvious way to achieve it.
So my question is: what is the magic trick to get the Cartesian product of binary operators given by any group such that it satisfy above-mentioned requirement? It would be really great if you stick to the $(\mathbb{Z}, +)$ example. 
 A: First, let me explain how a one-element category can be viewed as a monoid, since I think this already gets to the source of the confusion: that elements of the structure correspond to morphisms in one-element category, not "elements" of the unique object.
So suppose I have a one-element category $C$ with unique object $x$. There is a natural monoid $M_C$ associated to this category:


*

*Elements of $M_C$ are morphisms in $C$ - that is, the underlying set of $M_C$ is $Hom_C(x, x)$.

*The monoid operation $*$ in $M_C$ is given by composition: for $f, g\in Hom_c(x, x)$ we define $f*g=fg$.
Note that the elements of the monoid are the morphisms of $C$, not the elements of $x$ in any sense (indeed we don't need to think of $x$ as a thing that even has elements).
The structure of $M_C$ now comes from how $Hom_C(x, x)$ behaves.

OK, now I claim that every monoid arises in this way: that is, for a given monoid $A$, there is a category $C_A$ with a single object $x$ such that $M_{C_A}\cong M$.
It doesn't matter what our unique object is - let's take $x$ to be the number $37$. Now the hom-set $Hom_C(x, x)$ will be the set of elements of $A$, and the composition law will be given by $fg=f*g$ where $*$ is the monoid operation in $A$.
This feels totally artificial, and it is - but that's fine! A category is any structure satisfying [list of properties], and it's easy to check that the above structure does in fact satisfy those properties. In particular, the identity element $e_A$ of the monoid is the identity morphism $id_x$.
