Lemma 2. Let $E$ be a monoid and $x\in E$.
(1) There exists a unique homomorphism $f$ of $\mathbb{N}$ into $E$ with $f(1)=x$ and $$f(n)=\top^{n}x\text{, for all n in }\mathbb{N},$$ where $\top$ denotes the law of E.
(2) If $x$ is invertible, there exists a unique homomorphism $g$ of $\mathbb{Z}$ into $E$ such that $g(1)=x$ and $g$ coincides with $f$ on $\mathbb{N}$.
... We shall apply Lemma 2 to the case where the monoid $E$ is $\mathbb{Z}$; for every $m\in\mathbb{Z}$ there exists an endomorphism of $\mathbb{Z}$ characterized by $f_{m}(1)=m$. If $m\in\mathbb{N}$, the mapping $n\mapsto mn$ is an endomorphism of the magma $\mathbb{N}$; hence $f_m(n)=mn$ for all $m,n$ in $\mathbb{N}$. Multiplication on $\mathbb{N}$ can therefore be extended to multiplication on $\mathbb{Z}$ by the formula $mn=f_{m}(n)$ for $m,n\in\mathbb{Z}$. [emphasis added]
(Bourbaki Algebra Chapter 1, § 2, no. 6)
What is going on here? Can someone please explain this notion of multiplication with a little more detail? For example, what does the proof of commutativity of multiplication look like in this case? On the other hand, how can one, in general, extend an operation to larger domain "by a formula"?
In the part of the extract that you have emphasized, the word "definition" could easily have replaced the word "formula" - which has no special significance that I can see.
I don't have Bourbaki's book, but here is a way in which multiplication in $\mathbb{Z}$ can quickly be proved commutative using ideas similar to those in the quoted passage:
Use the same idea and the same notation generally when $E$ is an Abelian group in additive notation. (We could keep $E = \mathbb{Z}$ throughout the argument, if you prefer.) That is, for $x \in E$, let $f_x$ be the unique homomorphism $\mathbb{Z} \to E$ such that $f_x(1) = x$, and define $xn = f_x(n)$.
(Of course, this element is usually written as $nx$, or as $x^n$ when multiplicative notation is being used for $E$. Indeed, having defined multiplication in $\mathbb{Z}$, we can go on to define exponentiation in essentially the same way.)
If $x, y \in E$, then $f_x + f_y$ is a homomorphism $\mathbb{Z} \to E$, and $(f_x + f_y)(1) = x + y$, therefore $f_x + f_y = f_{x+y}$.
The zero homomorphism $z \colon \mathbb{Z} \to E$ satisfies $z(1) = 0$, therefore $f_0 = z$.
Let $n \in \mathbb{Z}$. By the previous two results, we have \begin{gather*} (x + y)n = xn + yn \quad (x, y \in E), \\ 0n = 0. \end{gather*} That is, the function $h_n \colon E \to E$, $x \mapsto xn$ is a homomorphism.
Take $E = \mathbb{Z}$; then $h_n$ is a homomorphism $\mathbb{Z} \to \mathbb{Z}$.
The identity homomorphism $i \colon \mathbb{Z} \to \mathbb{Z}$ satisfies $i(1) = 1$, therefore $f_1 = i$, therefore $h_n(1) = 1n = n$. But $f_n$ is the only homomorphism $\mathbb{Z} \to \mathbb{Z}$ such that $f_n(1) = n$. Therefore, $h_n = f_n$. That is, $mn = nm$ for all $m, n \in \mathbb{Z}$.