$ax\equiv b\pmod{\! n}$ has a unique solution if $\gcd(a,n) = 1$ [closed]

Yes, Bezout plays a key role, see below.

Theorem $\ $ The following are equivalent for integers $\rm\:c,\, m.$

$(1)\rm\ \ \ gcd(c,m) = 1$
$(2)\rm\ \ \ c\:$ is invertible $\rm\,(mod\ m)$
$(3)\rm\ \ \ x\to cx+d\:$ is $\:1$-$1\:$ $\rm\,(mod\ m)$
$(4)\rm\ \ \ x\to cx+d\:$ is onto $\rm\,(mod\ m)$

Proof $\ (1\Rightarrow 2)\ $ By Bezout $\rm\, gcd(c,m)\! =\! 1\Rightarrow cd\!+\!km =\! 1\,$ for $\rm\,d,k\in\Bbb Z\,$ $\rm\Rightarrow cd\equiv 1\!\pmod{\! m}$
$(2\Rightarrow 3)\ \ \ \rm cx\!+\!d \equiv cy\!+\!d\,\Rightarrow\,c(x\!-\!y)\equiv 0\,\Rightarrow\,x\!-\!y\equiv 0\,$ by multiplying by $\rm\,c^{-1}$
$(3\Rightarrow 4)\ \ $ Every $1$-$1$ function on a finite set is onto (pigeonhole).
$(4\Rightarrow 1)\ \ \ \rm x\to cx\,$ is onto, so $\rm\,cd\equiv 1,\,$ some $\rm\,d,\,$ i.e. $\rm\, cd+km = 1,\,$ some $\rm\,k,\,$ so $\rm\gcd(c,m)=1$

See here for a conceptual proof of said Bezout identity for the gcd.