This is a very elementary question, so please forgive me, but precisely because of this reason I haven't found an answer elsewhere:
How can it be proven that, among the integers, only $1$ and $-1$ have an integer multiplicative inverse? Or is that statement taken as an axiom?
On
Well, let's assume $ab=1$ when $a,b\in\mathbb{Z}$. Of course it means that both $a$ and $b$ are non zero. Because they are integers we can conclude that their absolute values are at least $1$. And then $1=|ab|=|a||b|\geq |a|$, and in the same way $1\geq |b|$. So both $a$ and $b$ must belong to the set $\{-1,1\}$.
On
It follows from the well-ordering principle for the natural numbers.
Suppose $ab=1$ for $a,b \in \Bbb{N}$ with $b>1$. Then $c=a(b-1)$ is a natural number with $0<c<1$. So $\{c,c^2,c^3,\dots\}$ is a set of natural numbers which has no least element, contradicting the well-ordering principle.
Now, if $ab=1$ for $a,b \in \Bbb{Z}$, then either $a,b$ or $-a,-b$ will fit into the proof in the above paragraph...
On
As we are interested in 1/a which is just a symbol being adopted to denote the inverse of an element in Z. Further identity element under multiplication in Z is 1. So if any member say ''a' of Z is invertible then there must be a member in Z denoted as 1/a or a^(-1) such that a. a^(-1) =1. Now it's a simple algebra that in Z 1/a is an integer iff a= 1 or -1. Hope it will help you.
On
It depends upon the way that $\mathbb Z$ and $\times$ are defined, of course, but I doubt that anyone ever created axioms for the integers such that what you stated is an axiom.
Suppos, for instance, that you define:
Then your statement becomes a proposition, which can in fact be proved.
On
First of all, of course integers, have multiplicative inverses and of course that statement is obviously false as $a = \frac 12, \frac 13, \frac 14,$ etc. bear out.
What you mean to say is $1$ and $-1$ are the only two integers whose multiplicative inverses are also integers.
I.e. If $a, \frac 1a \in \mathbb Z$ then $a = 1$ or $a = -1$.
That is not an axiom.
If $a \in \mathbb Z$ then you have five possibilities. $a > 1$ or $a = 1$ or $a = 0$ or $a = -1$ or $a < 1$.
Case 1: $a > 1$. Then if $\frac 1a < 0$ we would have $a*\frac 1a < a*0$ and $1 < 0$ which is a contradiction. We can't have $\frac 1a = 0$ because $0$ has no multiplicative inverses so $\frac 1a > 0$. If $\frac 1a \ge 1$ then $a*\frac 1a \ge a*1 = a$ or $1 \ge a$ which is a contradiction so $0< \frac 1a < 1$ and not an integer.
Case 2: $a = 1$. then $\frac 1a = 1$.
Case 3: $a = 0$ then $\frac 1a$ is undefined.
Case 4: $a = -1$ then $\frac 1a = -1$.
Case 5: $a < -1$. that is so similar to case 1: I'll leave it to you.
This is readily proved once you properly define $\Bbb Z$ and its operations. We may want to start from $\Bbb N$ with its successor map $S$ and the Peano axioms, then define addition $\Bbb N\times \Bbb N\to\Bbb N$ recursively as $x+0:=x$, $x+Sy:=S(x+y)$, and then define multiplication recursively as $x\cdot 0=0$, $x\cdot Sy:=x\cdot y+y$.
Now we see that $(a,b)\sim (c,d):\iff a+d=b+c$ is an equivalence relation on $\Bbb N^2$, define $\Bbb Z:=\Bbb N^2/{\sim}$ as the set of equivalence classes, define $\overline{(a,b)}+\overline{(c,d)}:=\overline{(a+c,b+d)}$ and define $\overline{(a,b)}\cdot \overline{(c,d)}:=\overline{(a\cdot c+b\cdot d, a\cdot d+b\cdot c)}$ (which involves showing that this is well-defined). The claim now boils down to $\overline{(a,b)}\cdot\overline{(c,d)}=\overline{(1,0)}\implies a=Sb\lor b=Sa$. It does take a bit of work, but by going back the way we came can ultimately be reduced to a statement about $\Bbb N$ that is provable by induction.