Deriving an inverse Mobius function

The book claims that the inverse of a Mobius function is another Mobius function of the form $f^{-1}(z)=\frac{dz-b}{-cz+a}$. I am having trouble with combining the individual inverse functions to find this form.
So starting with breaking down a Mobius function to its individual actions:
$$\begin{aligned}
&\frac{az+b}{cz+d}\\
&=\frac{a(z+\frac{d}{c}-\frac{d}{c})+b}{c(z+\frac{d}{c}}\\
&=\frac{a}{c}+\frac{b-a\frac{d}{c}}{c(z+\frac{d}{c}}\\
&=\frac{a}{c}+\frac{bc-ad}{c^2}*\frac{1}{z+\frac{d}{c}}\\\end{aligned}$$
Thus the individual functions and their inverses are:
$$\begin{aligned}
&f(z_1)= z+\frac{d}{c} \quad \quad f^{-1}(z_1)= z-\frac{d}{c} \\
&f(z_2)=\frac{1}{z} \quad \quad f^{-1}(z_2)=\frac{1}{z}\\
&f(z_3)= z*\frac{bc-ad}{c^2} \quad \quad f^{-1}(z_3)= z*\frac{c^2}{bc-ad}\\
&f(z_4)=z+\frac{a}{c} \quad \quad f^{-1}(z_4)=z-\frac{a}{c}\end{aligned}$$

Which I combined into the form:
$$\begin{aligned}
&\frac{1}{z-\frac{d}{c}}*\frac{c^2}{bc-ad}-\frac{a}{c}\\
&=\frac{c^2}{(z-\frac{d}{c})(bc-ad)}-\frac{a}{c}\end{aligned}$$

At this point I hit a wall and gave up after trying a bunch of different algebraic techniques. I can't figure out how to simplify to the inverse given by the book. Anyone have any ideas? It might just be simple algebra but it has me confused.

I am a little confused why you are breaking the function down into its separate pieces. If we solve a mobius transformation for $z$, we get
\begin{align*}
&f(z)=\frac{az+b}{cz+d}\\
\Rightarrow&f(z)(cz+d)=az+b\\
\Rightarrow&czf(z)+f(z)d=az+b\\
\Rightarrow&f(z)d-b=az-czf(z)\\
\Rightarrow&f(z)d-b=z(a-cf(z))\\
\Rightarrow&z=\frac{f(z)d-b}{a-cf(z)}\\
\Rightarrow&f^{-1}(z)=\frac{dz-b}{cz-a}
\end{align*}
That is how I solved the problem. I am not sure if there is another method.

@DPR The decomposition that you are using is great to show that any Mobius transformation is a composition of a few simple operations. While it's great for that purpose, I agree with @cdunn that the simpler algebraic definition of a Mobius transformation is more natural for what you want to accomplish. An alternative to solving the equation $w=(az+b)/(cz+d)$ is to simply plug the proposed inverse into the the original function:

$$\begin{align}
\frac{a\frac{dz-b}{-cz+a}+b}{c\frac{dz-b}{-cz+a}+d} &=
\frac{(adz-ab-bcz+ab)/(-cz+a)}{(cdz-bc-cdz+ad)/(-cz+a)} \\
&= \frac{(ad-bc)z}{ad-bc}= z.
\end{align}
$$

Let $z_0\in\mathbb{C}$. Define $f(z_0)=\frac{az_0+b}{cz_0+d}=z_1$. Note this function receives $z_0$ and returns $z_1$. Thus, the inverse function $f^{-1}$ should receive $z_1$ and return $z_0$. Solving $f(z_0)=z_1$ for $z_0$ we find

$$\begin{align}&\frac{az_0+b}{cz_0+d}=z_1\\ \implies&az_0+b=z_1(cz_0+d)\\ \implies&(a-z_1c)z_0=z_1d-b\\ \implies&z_0=\frac{z_1d-b}{a-z_1c}=f^{-1}(z_1)\end{align}$$

which is precisely the inverse function we have been looking for!