Edit: I realized that, in your question, the Fourier series decomposition of the temperature should go as (please confirm the braces after $n$) \begin{equation} T(t) = T_0 + \sum_{n=1}^\infty T_n e^{in (\omega t - \phi)} \end{equation}
Original answer: The only equation that contains any physical link between $T(t)$ and $S(t)$ is the one you have described: \begin{equation} \sigma T^4(t) = (1-A)S(t). \end{equation} Taking the time derivative and then dividing both sides by $\sigma T^4(t)$, you get \begin{align} 4\frac{\sigma T^3(t)}{\sigma T^4(t)} \frac{dT(t)}{dt} &= \frac{(1-A)}{\sigma T^4(t)} \frac{dS(t)}{dt}\\ & \\ \Rightarrow S(t) \frac{d T(t)}{dt} &= \frac{1}{4} T(t) \frac{dS(t)}{dt}, \end{align} where in the second equation, I have replaced $\sigma T^4(t)$ using the above equation and simplified further. Here, you can substitute the respective Fourier expansions and obtain \begin{align} S_0\left ( 1 + \sum_{m=1}^\infty \frac{S_m}{S_0} e^{im\omega t} \right ) \left( \sum_{n=1}^\infty T_n (i n \omega) e^{in(\omega t - \phi)} \right ) = \frac{1}{4}T_0 \left( 1 + \sum_{n=1}^\infty \frac{T_n}{T_0} e^{i n(\omega t - \phi)}\right) \\ \times \left( \sum_{m=1}^\infty S_m (i m\omega)e^{i m \omega t}\right ) \end{align}
\begin{align} \Rightarrow \left ( 1 + \sum_{m=1}^\infty \frac{S_m}{S_0} e^{im\omega t} \right ) \left( \sum_{n=1}^\infty \frac{T_n}{T_0} (i n \omega) e^{in(\omega t - \phi)} \right ) = \frac{1}{4} \left( 1 + \sum_{n=1}^\infty \frac{T_n}{T_0} e^{i n(\omega t - \phi)}\right) \\ \times \left( \sum_{m=1}^\infty \frac{S_m}{S_0} (i m \omega) e^{i m \omega t}\right ) \end{align} which, upon re-ordering the terms and comparing the coefficients for each unique Fourier mode, should give you the required result.