Solution

Our capacitor can be represented as two capacitors connected in series; for them the capacitance is
$$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}\tag{1}$$
where
$$C_i=\frac{\varepsilon_i\varepsilon_0S}{d_i}\tag{2}$$
$(2)\to(1):$
$$\frac{1}{C}=\frac{1}{\varepsilon_0S}\left(\frac{d_1}{\varepsilon_1}+\frac{d_2}{\varepsilon_2}\right)$$
$$C=\frac{\varepsilon_0\varepsilon_1\varepsilon_2S}{\varepsilon_2d_1+\varepsilon_1d_2}\tag{3}$$

Now recall that at the interface between dielectrics the normal component of the field undergoes a jump. The dielectrics are isotropic, the field is normal to the interface, and there are no free charges at the interface, so
$$\varepsilon_2E_2=\varepsilon_1E_1\tag{4}$$
For the voltage across the capacitor after placing the charge:
$$V=E_1d_1+E_2d_2=\frac{q}{C}\tag{5}$$
The surface density of bound (induced) charges at the interface is determined by the jump in the normal component of the polarisation vector:
$$\sigma'=P_{n1}-P_{n2}\tag{6}$$
For an isotropic dielectric
$$\vec P=(\varepsilon-1)\varepsilon_0\vec E$$
Therefore
$$\sigma'=(\varepsilon_1-1)\varepsilon_0E_1-(\varepsilon_2-1)\varepsilon_0E_2 =\varepsilon_0[(\varepsilon_1E_1-\varepsilon_2E_2)+E_2-E_1]$$
Taking $(4)$ into account, we finally obtain a nice expression:
$$\sigma'=\varepsilon_0(E_2-E_1)\tag{7}$$

Solve the system $(4), (5), (7)$:
$$E_1=\frac{q\varepsilon_2}{C(d_1\varepsilon_2+d_2\varepsilon_1)}, \quad E_2=\frac{\varepsilon_1}{\varepsilon_2}E_1=\frac{q\varepsilon_1}{C(d_1\varepsilon_2+d_2\varepsilon_1)}$$
$$\sigma'=\varepsilon_0q\frac{\varepsilon_1-\varepsilon_2}{C(d_1\varepsilon_2+d_2\varepsilon_1)} =\varepsilon_0q\frac{\varepsilon_2d_1+\varepsilon_1d_2}{\varepsilon_0\varepsilon_1\varepsilon_2S}\, \frac{\varepsilon_1-\varepsilon_2}{d_1\varepsilon_2+d_2\varepsilon_1}$$
$$\sigma'=\frac{q'}{S}=\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1\varepsilon_2}\frac{q}{S}$$
$$q'=\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1\varepsilon_2}q$$