Proof

For a given real number $x\in\mathbb R$, $x\neq 0$, let $y$ be any real number with $xy=1\label{E18322}\tag{1}.$
Applying the existence of inverse real numbers with respect to multiplication, we can multiply both sides of the equation by $x^{-1}$ from the left side, resulting in $x^{-1}(xy)=x^{-1}\cdot 1.$
From the existence of real one and since the multiplication of real numbers is cancellative, it follows that $x^{-1}(xy)=x^{-1}.$
Because the multiplication of real numbers is associative, it follows $(x^{-1}x)y=x^{-1}.$
Applying the existence of inverse real numbers with respect to multiplication once again leads to $1\cdot y=x^{-1}.$
Using the existence of real one again, results in $y=x^{-1}.$
Thus, the there is only one such real number $y$ for which $x\cdot y=1$, namely $y=x^{-1}$.
Note that because the multiplication of real numbers is commutative, we would get the same result, if we multiplied both sides of the equation $\eqref{E18322}$ from the right side by $x^{-1}$.
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Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983