Here is the answer you have provided.
It lost me after the first sentence ...
I don't understand that answer, but algebraically it works out easily enough.
We need
\(\displaystyle \frac{a(ax+b)/(cx+d)+b}{c(ax+b)/(cx+d)+d}\equiv x.\)
Clearing the fraction within the fraction on the lhs,
\(\displaystyle \frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)}\equiv x. \)
For this to work we need three things to happen.
(1) the constant term on the top line should be zero, ie. ab + bd = b(a + d) = 0,
(2) the x coefficient on the bottom line should be zero, ie. ca + dc = c(a + d) = 0,
(3) The resulting fraction should cancel down to leave just x.
Applying (1) and (2) reduces the lhs to
\(\displaystyle \frac{x(a^{2}+bc)}{cb+d^{2}}\)
and this cancels down if
\(\displaystyle a^{2}=d^{2}.\)
\(\displaystyle a+d=0,( \text{ so }a=-d),\) meets all of the requirements.