Suppose $ab=1$. In particular $a,b \neq 0$. Note that you have $|ab| \geq |a|$ and $|ab| \geq |b|$ since we are working with integers (you can prove easily by induction on $m$ that for all $n,m \in \mathbb{N}\setminus \{ 0 \} $ you have $ nm \geq n$).
So $1 = |ab| \geq |a|$. This means that $a \in \{ -1, 0, 1\}$.
This means that $\mathbb{Z}^* \subseteq \{ -1, 1\}$. Clearly, the opposite inclusion is trivial.
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If $x$ is a unit $\mathbb{Z}$, then there is one $y \in \mathbb{Z}$ such that $xy = 1.$ So finding the units in $\mathbb{Z}$ is the same as finding the integer solutions of the equation $xy = 1.$
Suppose $ab=1$. In particular $a,b \neq 0$. Note that you have $|ab| \geq |a|$ and $|ab| \geq |b|$ since we are working with integers (you can prove easily by induction on $m$ that for all $n,m \in \mathbb{N}\setminus \{ 0 \} $ you have $ nm \geq n$).
So $1 = |ab| \geq |a|$. This means that $a \in \{ -1, 0, 1\}$.
This means that $\mathbb{Z}^* \subseteq \{ -1, 1\}$. Clearly, the opposite inclusion is trivial.