Will $h(x)=\frac{ax+b}{cx+d}$, where $c\neq0$, always satisfy $h(h\cdots(h(x))\cdots)=x$ in a finite number of iterations? Consider function $f(x) = \frac{x-5}{x-3}$ , after four iterations we arrived at that $f(f(f(f(x))))) = x$ , likewise similarily for $g(x) = \frac{1}{1-x}$ we have $g(g(g(x))) = x$.
So, is it always possible that the function of form
$$h(x) = \frac{ax+b}{cx+d}$$ where $c \neq 0$ will satisfy
$h(h\cdots(h(x))\cdots) = x$ in a finite number of iterations?
 A: 
is it always possible that the expression of form $h(x)$ = $\dfrac{ax+b}{cx+d}$ where $c \neq 0$ will satisfy $h(h\cdots (h(x))\cdots) = x$

Let's use the following notation:  For a 2×2 matrix
$$M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ denote its action on a (real or complex) number as
$$Mz = \frac{az+b}{cz+d} \tag1$$
so that $M$ is a linear fractional transformation$^1$. One propery is that the action is transitive, which means that for two such transformations $M_1$ and $M_2$ there is the relation
$$(M_2\circ M_1)(z) = M_2 (M_1(z)) = (M_2M_1)(z)$$
where $M_2M_1$ denotes the usual matrix product, and $\circ$ denotes composition of functions. This means that for applying a transformation $n$ times, we have
$$(\underbrace{M\cdots M}_{n\text{ times}}) (z) = (\underbrace{M\circ\cdots\circ M}_{n\text{ times}})(z)$$
Therefore, investigating iterated fractional linear transforms is basically the same like investigating powers of matrices. In  order for the $n$-th iterate of $M$ to come out as identity, the $n$-th power of $M$ must satisfy
$$M^n = \begin{pmatrix}w&0\\0&w\end{pmatrix},\qquad w\neq0 \tag2 $$
This can be stated as: The eigenvalues of $M^n$ are $w$.  This means if we find a matrix $M$ whose eigenvalues are both $n$-th roots of $w$ and $M$ is diagonalizable, then $M^n$ satisfies the condition above.
$\def\tM#1#2#3#4{{\big(\begin{smallmatrix}{#1}&{#2}\\{#3}&{#4}\end{smallmatrix}\big)}}$
Examples
For example, your first transform $M_f=\tM{1}{-5}{1}{-3}$ has characteristic polynomial $\lambda^2+2\lambda +2$ with zeros $\lambda=-1\pm i$. Now $\lambda^2=\mp2i$ and thus $\lambda^4 = -4$ so that $M_f$ satisfies $(2)$ with $n=4, w=-4$.
Your second transform $M_h=\tM{0}{1}{-1}{1}$ has characteristic polynomial $\lambda^2-\lambda +1$ with roots $\lambda=(1\pm\sqrt{-3})/2$ that are 3rd roots of $-1$, i.e. $M_h$  satisfies $(2)$ with $n=3, w=-1$.
The function $z\mapsto1/(1+z)$ from the comment has $M_c=\tM{0}{1}{1}{1}$ with characteristic polynomial $\lambda^2-\lambda-1$ and zeros $\phi$ and $-1/\phi$ where $\phi$ is the golden ratio. $\phi^n = (-1/\phi)^n$ implies $\phi^{2n} = (-1)^n$ which has only the solution $n=0$ because $|\phi|\neq1$.
In General
In general, the condition $M^n=wE$ does not hold for any $n$, which means your conjecture about iterates does not hold$^2$.  The set for which it holds is a set of measure zero.  In particular, the quotient of the eigenvalues must be some root of unity.  For example, you could start with two eigenvalues like $\lambda_i=\zeta_5^i$ where $\zeta_5$ is a primitive 5-th root of unity and from there construct a matrix with that eigenvalues.  Then the matrix satisfies $(2)$ with $n=5$ and $w=1$.
Starting with a random matrix, however, the quotient of the eigenvalues will in general not equal some root of unity.
Notice that the condition for the eigenvalues of $M$ are necessary, but not sufficient for $M^n=wE$.  For example, $M=\tM{1}{0}{1}{1}$ has 1 as eigenvalues, but $M^n=\tM{1}{0}{n}{1}\neq wE$ for any $n\neq 0$. What's also needed is that $M$ is diagonalizable.
Notes
Also notice that even in the case $M:\Bbb R\to\Bbb R$ the analysis might involve complex numbers, like in your examples where both $M_f$ and $M_h$ have complex eigenvalues.
Moreover, when you found some matrix $M$ that satisfies $(2)$, you can construct new transformations $M'=AMA^{-1}$ where $A$ is a 2×2 matrix with $\det A\neq 0$:
$$M'^{\,n} = (AMA^{-1})^n = AM^nA^{-1} \stackrel{(2)}= A(wE)A^{-1} = wE$$
where $E$ denotes the 2×2 identity matrix.  This means $M'$ satisfies $(2)$ again.  And representations of fractional linear transforms by means of $(1)$ are not unique because $M$ and $\alpha M$ represent the same function provided $\alpha\neq 0$.

$^1$One usually demands that $\det M = ad-bc\neq 0$ so that the function is not constant.  And then the inverse of $M$ exisits and is represented by the inverse matrix $M^{-1}$.
$^2$Except for the trivial case $n=0$ because $M^0=E$ per definition.  But $M^0$ means "apply $M$ zero times", so we don't consider that case.
