So you're essentially switching the big-O and the expectation, i.e., $E[O((S_n -1)^2)] = O(E[(S_n -1)^2]) = O(1/n)$. Is this always valid, or do we need some extra assumptions for this step?

How does the asymptotic growth of the second term follow? E.g., why wouldn't this be $O(1/\sqrt{n})$ or $O(1/\log{n})$ or anything else?

The second order term in the Taylor expansion should be $O((S_n - 1)^2)$ of course.

Right, this seems to yield something similar to the expression in terms of the variance I have above. Taking expectations on the Taylor expansion, I'd get $E[\sqrt{Sn}] = 1 + E[O(S_n - 1)]$. I'm not sure what to make of that second term.

Very nice! I'll try to go through a similar approach for the general case with $\sigma \neq 1$ (or does your approach crucially rely on $\sigma=1$ somewhere?). It also seems that the Berry--Esseen inequality should let us get an explicit (rather than asymptotic) result involving $\mu, \sigma$ and $1/\sqrt{n}$, right?