How to derive the (dry) deposition velocity for particles in the atmosphere
After some more work I got it working, this might not be the shortest solution possible. There are no further assumptions needed, just pure algebra.
The solution is a bit tedious to reach, so it's not too surprising that this is usually not included in the literature.

Start by splitting $(3)$ into two separate equations
$$
\begin{cases}
\frac{C_3}{r_t}=\frac{C_3-C_2}{r_a}+v_g C_3\hspace{10mm}(4)\\
\frac{C_3}{r_t}=\frac{C_2}{r_b}+v_g C_2\hspace{17mm}(5)
\end{cases}
$$
Get $C_3$ by multiplying $(5)$ with $r_t$ and breaking out $C_2$:
$$
C_3 = C_2 r_t (\frac{1}{r_b} + v_g)\hspace{10mm}(6)
$$
Divide $(4)$ by $C_3$:
$$
\frac{1}{r_t}=\frac{1-C_2/C_3}{r_a}+v_g\hspace{10mm}(7)
$$
Insert $(6)$ in $(7)$:
$$
\frac{1}{r_t}
=\frac{1-\frac{C_2}{C_2r_t(1/r_t+v_g)}}{r_a}+v_g 
=\frac{ \frac{1}{r_b}+v_g-\frac{1}{r_t} }{ r_a(\frac{1}{r_b}+v_g)}+v_g
\hspace{10mm}
$$
Move $r_t$ to LHS, i.e. add $\left(r_t \cdot r_a(\frac{1}{r_b}+v_g)\right)^{-1}$ to both sides:
$$
\frac{1}{r_t}\left(1 + \frac{1}{r_a(\frac{1}{r_b}+v_g)}\right)
= \frac{1}{r_a}+v_g
$$
Divide by $\left(1+\frac{1}{r_a(\frac{1}{r_b}+v_g)} \right)$ on both sides:
$$
\frac{1}{r_t}
=\frac{1/r_a+v_g}{\left(1+\frac{1}{r_a(\frac{1}{r_b}+v_g)} \right)}
$$
Multiply numerator and denominator (RHS) by $r_a$ and simplify:
$$
= \frac{1+v_g r_a}{\left(r_a+\frac{1}{(\frac{1}{r_b}+v_g)} \right)}
= \frac{1+v_g r_a}{\left(r_a\frac{(\frac{1}{r_b}+v_g)}{(\frac{1}{r_b}+v_g)}+\frac{1}{(\frac{1}{r_b}+v_g)} \right)}
= \frac{(1+v_gr_a)(\frac{1}{r_b}+v_g)}{r_a(\frac{1}{r_b}+v_g)+1}
$$
Re-organize the numerator and split it:
$$
=\frac{ \frac{1}{r_b}+v_g+v_g\frac{r_a}{r_b} +v_g^2 r_a }{r_a(\frac{1}{r_b}+v_g)+1}
=\frac{ \frac{1}{r_b}+v_g\left(r_a(\frac{1}{r_b} +v_g)+1\right) }{r_a(\frac{1}{r_b}+v_g)+1}
=\frac{ \frac{1}{r_b}}{r_a(\frac{1}{r_b}+v_g)+1} + v_g
$$
If we now move $r_b$ in the first term on the RHS to the denominator we end up with the final expression:
$$
\frac{1}{r_t} = \frac{1}{r_a + r_b + r_a r_b v_g } + v_g
$$