Can I just take the resistances in series and the capacitances in parallel?
That would not be possible as \$C_1\$ and \$C_2\$ are not in parallel and \$G\$ and \$R_s\$ are not in series.
The most "simple" way would be to use regular parallel/series formula's to calculate the total equivalent impedance.
$$R_{series} = R_1 + R_2$$
$$R_{parallel} = R_1//R_2 = \frac{R_1R_2}{R_1+R_2}$$
So this would give you something along the lines of
$$\begin{align}
Z &= (\frac{1}{G}//\frac{1}{C_1s} + R_s)//\frac{1}{C_2s}\\
&= \left[\left(\frac{1}{G+C_1s} + R_s\right)^{-1} + C_2s\right]^{-1}
\end{align}$$
Alternatively, you can also use 2-EET (extra-element theorem) to get the time constants. Let's say \$G = \frac{1}{R}\$.
$$Z = Z^{(dc)}\frac{1 + \frac{Z_{n1}}{Z_1} + \frac{Z_{n2}}{Z_2} + \frac{Z_{n1}}{Z_1}\frac{Z_{n2}^{(1)}}{Z_2}}{1 + \frac{Z_{d1}}{Z_1} + \frac{Z_{d2}}{Z_2} + \frac{Z_{d1}}{Z_1}\frac{Z_{d2}^{(1)}}{Z_2}}$$
Where each of these terms is actually quite easy to find if you know the method (\$Z_1 = \frac{1}{C_1s}\$ and \$Z_2 = \frac{1}{C_2s}\$). We find
$$\begin{align}
Z^{(dc)} &= R + R_s\\
Z_{n1} &= R//R_s\\
Z_{n2} &= 0\\
Z_{n2}^{(1)} &= 0\\
Z_{d1} &= R\\
Z_{d2} &= R + R_s\\
Z_{d2}^{(1)} &= R_s
\end{align}$$
So plugging these in you get
$$\begin{align}
Z &= \left(R + R_s\right)\frac{1 + (R//R_s)C_1s}{1 + RC_1s + (R+R_s)C_2s + RR_sC_1C_2s^2}\\
&= \frac{A}{1 + \tau_1s} + \frac{B}{1 + \tau_2s}
\end{align}$$
You have two time constants in this circuit at the roots of the denominator. If these time constants are expected to be far apart, you can use a common approximation:
$$\begin{align}
d(s) &= 1 + as + bs^2\\
\tau_1 &\approx a\\
\tau_2 &\approx \frac{b}{a}
\end{align}$$
(it is an assumption that if \$s\$ is small then \$s^2\$ is much smaller. If that makes \$as \gg bs^2\$ then \$bs^2\$ is negligible).
You can then approximate the dominant (slowest) time-constant using
$$\tau_1 \approx RC_1 + (R+R_s)C_2$$
You should always verify by approximating \$\tau_2\$ as well and then checking that it is much smaller. If it isn't you'll have to solve the quadratic equation.
