Here is a "slightly" more general approach.
Define $N:\Bbb Z \to \Bbb N$, by $N(k) = |k|$ (the absolute value).
You can prove that $N(km) = N(k)N(m)$ (consider the $4$ cases:
1) $k,m \geq 0$
2) $k \geq 0, m < 0$
3) $k < 0, m \geq 0$
4) $k,m < 0$ separately).
An invertible (multiplicative) element $x \in \Bbb Z$ is one for which there is a $y \in \Bbb Z$, with $xy = 1$.
Thus: $N(x)N(y) = N(xy) = N(1) = 1$.
Now if $N(x) > 1$ (which is just saying $|x| > 1$), but $N(x)N(y) = 1$, then:
$1 = N(x)N(y) > N(y)$.
But the ONLY natural number less than $1$ is $0$, so that $y = 0$ and $N(0) = 0$, so:
$1 = N(x)N(y) = N(x)N(0) = N(x)0 = 0$, a contradiction.
If $N(x) = 0$, then $x = 0$, and $N(0)N(y) = 0$ (no matter what $y$ we pick), so $N(x)N(y) \neq 1$,
and so $0$ is not invertible.
That means if $x$ is invertible, we must have $N(x) = 1$. Which integers have $N(x) = 1$?
The only two are $1$ and $-1$, and we can check directly these two are invertible:
$(1)(1) = 1$ and $(-1)(-1) = 1$.
Now, see if you can extend this to $\Bbb Z \times \Bbb Z$ by showing that:
$(a,b)$ is invertible in $\Bbb Z \times \Bbb Z$ if and only if $a,b$ are both invertible in $\Bbb Z$ (recall that the multiplicative identity of $\Bbb Z \times \Bbb Z$ is $(1,1)$, and we multiply "coordinate-wise").
There is a method to my madness-for "more complicated" structures you will meet later, there are some similar $N$ functions, and you will get more mileage out of this approach.