The hypothesis simply says that for every $a$, there is at most one function $f$ of the form $ax+b$ in our group ($G$ is a group). On the other hand, it's clear that the conclusion we want to reach is that if $f_a\equiv ax+b,\ f_c\equiv cx+d,\ a,c\ne 1$, then $\frac{b}{a-1}=\frac{c}{d-1}$. Suppose the contrary. Then $f_af_c\equiv acx+ad+b,\ f_cf_a=cax+bc+d$. By assumption, we have $ad+b\ne bc+d$, which means that $f_af_c,f_cf_a$ are two distinct functions in our group with leading coefficient $ac$, and this contradicts the hypothesis.

Let $f_1=ax+b,f_2=ax+c$ Then $f^{-1}_1=\frac{x-b}{a} \in G$ $g=f_2(f^{-1}_1)=a \frac{x-b}{a}+c=x-b+c \in G$ But $g(x)=x$ has solution only for $b=c$

Solution from Twitch Solves ISL: Note that the only element of $G$ of the form $x \mapsto x+\lambda$ is the identity. Now consider any two non-identity functions of $G$, say \begin{align*} f(x) &= ax+b \\ g(x) &= cx+d \end{align*}so that $a \neq 1$ and $c \neq 1$. The (unique) fixed point of $ax+b$ is $\frac{b}{a-1}$ and the (unique) fixed point of $cx+d$ is $\frac{d}{c-1}$, so we just need to show \[ \frac{b}{a-1} = \frac{d}{c-1}. \]Note the inverses $f^{-1}(x) = \frac{x-b}{a}$ and $g^{-1}(x) = \frac{x-d}{c}$ should also be in $G$. Then consider the composition \begin{align*} g(f(g^{-1}(f^{-1}(x)))) &= c \cdot \left[ a \cdot \frac{\frac{x-b}{a}-d}{c} + b \right] + d \\ &= x - b - ad + bc + d. \end{align*}Since this is an element of $G$, as we commented before we must have \[ b - ad + bc + d = 0 \]which was what we wanted.