How can one prove the equation for a system of lenses? Let's say we have a system of 2 lenses which are placed in a non-negligible distance $L$ from eachother. 
How can I prove without using geometrical construction and principal planes that
$$\frac 1f = \frac {1}{f_1} + \frac {1}{f_2} -\frac{L}{f_1 f_2} $$
describes the system's focal length?

For example, with $ L \lt \lt f_1f_2 $ :
$\dfrac{1}{f_1} = \dfrac{1}{o_1} + \dfrac{1}{i_1}$ where $o_1$ is the object distance and $i_1$ the image distance of the first lens. 
And of course, $\dfrac{1}{f_2} = \dfrac{1}{o_2} + \dfrac{1}{i_2}$ .
Now you can say the image of the first lense is the object distance of the second length (with paying attention to conventions what is negative and what positive):
$o_2 = -i_1$ 
Using this relation one receives $\dfrac{1}{o_1} + \dfrac{1}{i_2} = \dfrac{1}{f_1} + \dfrac{1}{f_2}  = \dfrac{1}{f}$ 

I already tried saying $o_2 = -i_1 + L$ but this didn't lead anywhere.

Use ray transfer matrices. The system matrix is given by
$$
S=\begin{pmatrix}
1 & 0 \\
-\frac{1}{f_2} & 1 \\
\end{pmatrix}
\begin{pmatrix}
        1 & L \\
        0 & 1 \\
        \end{pmatrix}
\begin{pmatrix}
        1 & 0 \\
        -\frac{1}{f_1} & 1 \\
        \end{pmatrix}
$$
The back focal length $BFL$ is determined by the point where all the rays in an incoming collimated beam meet after the second lens:
$$
\begin{pmatrix}
        1 & BFL \\
        0 & 1 \\
        \end{pmatrix}
S
\begin{pmatrix}
        z \\
        0  \\
        \end{pmatrix}
=
\begin{pmatrix}
        0 \\
        \text{X}  \\
        \end{pmatrix}
$$
for all $z$. Solve for $BFL$ and then find $f$ by adding the distance of the second lens to the origin of coordinates. The origin is not arbitrary, it is chosen to a specific point on purpose in order to give a neat symmetric expression.
