= Solution
Independence and the normal exponential moment give $\mathbb E[S_t\mid\mathcal F_s]=S_s\exp[(\mu+\sigma^2/2)(t-s)]$. Thus
$$
\boxed{\mu=-\sigma^2/2.}
$$
For $\sigma\ne0$, $B_t/t\to0$ almost surely, so $\log S_t/t\to-\sigma^2/2$ and $S_t\to0$ almost surely. Nevertheless $\mathbb E S_t=e^{x_0}$ for every finite $t$, whereas $\mathbb E S_\infty=0$. This <Exponential martingale for Brownian motion> is not <uniformly integrable>, so it does not converge in $L^1$. If $\sigma=0$, the process is instead constantly $e^{x_0}$.
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