How to solve this integral equation with the Dirichlet kernel? It is 
$$  S (t) = 1  - i g \int_0^t d \tau \left( \sum_{n=-M}^M e^{-i n (t- \tau )} \right) S(\tau) . $$
The kernel is the Dirichlet kernel. Numerical result is shown in the figure. 
The $M\rightarrow \infty $ limit is easy (because the kernel reduces to a series of delta functions). In this limit, $S$ is a piece-wise constant function, illustrated by the red lines. But for a finite $M$, you will always get the fast rotation whose amplitude does not decay to zero in the limit of $M \rightarrow \infty $. Its amplitude converges to a finite value actually.  
The purpose is not to get the analytical expression of $S $ (it might does not exist), but to understand qualitatively its behavior.

 A: Nice solenoids...:)
What follows is not a full answer, but hopefully may help 


*

*1) Have you tried to take the Fourier Transform of LHS and RHS (the RHS is clearly a convolution) ? 

*2) As the Dirichlet kernel is a way to work on partial sums of Fourier series, when you say "converges to a finite value", I am almost sure that it has something to do with the Gibbs' phenomenon and its quantization (overshoot by $\approx 0.09 \%$) (https://en.wikipedia.org/wiki/Gibbs_phenomenon)

*3) Besides, have a look at "The good, the bad and the ugly of kernels: why the
Dirichlet kernel is not a good kernel"(https://www.google.fr/search?q=good+ugly+kernels+Dirichlet&ie=utf-8&oe=utf-8&client=firefox-b&gfe_rd=cr&ei=n74uV-TpDKus8weJgoS4DQ)
A: The kernel can be written
$$\sum_{n=-M}^M e^{in(t-\tau)}=\frac{\sin\left((M+1/2)(t-\tau)\right)}{\sin\left(\frac12(t-\tau)\right)}$$
The kernel has zeroes when $t-\tau=\frac{2\ell \pi}{2M+1}$, $\ell \in \mathbb{Z}$, $\ell \ne (2M+1)m$, for $m\in \mathbb{Z}$.  When $t-\tau\to 2m\pi$, we have
$$\lim_{t-\tau\to 2m\pi }\left(\frac{\sin\left((M+1/2)(t-\tau)\right)}{\sin\left(\frac12(t-\tau)\right)}\right)=2M+1$$
for $m\in \mathbb{Z}$. 
Next, note that the limit, $\lim_{M\to \infty}\left(\frac{\sin\left((M+1/2)(t-\tau)\right)}{\sin\left(\frac12(t-\tau)\right)}\right)$, does not exist.   However, we can evaluate
$$\lim_{M\to \infty} \int_0^t \sum_{n=-M}^M e^{in(t-\tau)}\,S(\tau)\,d\tau$$
using the approach I presented in THIS ANSWER.    Proceeding analogously, we have for $0\le t<2\pi$
$$\begin{align}
\int_0^t \sum_{n=-M}^M e^{in(t-\tau)}\,S(\tau)\,d\tau&=\int_0^t\frac{\sin\left((M+1/2)(t-\tau)\right)}{\sin\left(\frac12(t-\tau)\right)}S(\tau)\,d\tau\\\\
&=2\int_0^{(2M+1)t/2}\frac{\sin(x)}{(2M+1)\sin\left(\frac{x}{2M+1}\right)}S\left(t+\frac{2x}{2M+1}\right)\,dx\\\\
&\to \pi S(t)\,\,\text{as}\,\,M\to \infty \tag 1
\end{align}$$
Note that if $2k\pi <t<2(k+1)\pi$, then $(1)$ becomes
$$\lim_{M\to \infty}\int_0^t \sum_{n=-M}^M e^{in(t-\tau)}\,S(\tau)\,d\tau=\pi\left(\sum_{j=1}^k S(2j\pi)+S(t)\right)$$ 
