Put $f^k(x) = \underbrace{f(\dots(f(x))\dots)}_{k \text{ times}}$. Then we have $f^k(x) = \frac{a_k x + b_k}{c_k x + d_k}$. From
$f^{k+1}(x) = f(f^k(x))$ we get:
$$\begin{align*} a_{k+1} &= a a_k + b c_k \ b_{k+1} &= a b_k + b d_k \ c_{k+1} &= c a_k + d c_k \ d_{k+1} &= c b_k + d d_k \end{align*}$$
But we also have $f^{k+1}(x) = f^k(f(x))$, so
$$\begin{align*} a_{k+1} &= a a_k + c b_k \ b_{k+1} &= b a_k + d b_k \ c_{k+1} &= a c_k + c d_k \ d_{k+1} &= b c_k + d d_k \end{align*}$$
Comparing, we get $b c_k = c b_k$ and $(a-d) b_k = b(a_k - d_k)$.
If $f^n(0) = 0$, then $b_n = 0$, so $b c_n = 0$. We are given that $f(0) \neq 0$, so $b \neq 0$. Hence $c_n = 0$. Also $b(a_n - d_n) = 0$ and hence $a_n = d_n$. Hence $f^n(x) = x$.