Solution:

Composing, $f(f(x)) = \frac{(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)},$ and this equals $x$ identically only if $c(a + d) = 0.$ Since $c \ne 0,$ we get $d = -a,$ so $f(x) = \frac{ax + b}{cx - a}.$

A fixed point satisfies $cx^2 - ax = ax + b,$ i.e. $cx^2 - 2ax - b = 0,$ whose roots are $19$ and $97.$ By Vieta's formulas, $19 + 97 = \frac{2a}{c},$ so $\frac{a}{c} = 58.$

Finally, $y$ is in the range exactly when $y = \frac{ax + b}{cx - a}$ has a solution, i.e. $x(cy - a) = ay + b.$ This solves for $x$ unless $cy - a = 0;$ and when $y = \frac{a}{c},$ the right side $\frac{a^2}{c} + b = \frac{a^2 + bc}{c}$ is nonzero (otherwise $f$ would be constant). So the unique number not in the range is $\frac{a}{c} = 58.$