Do I understand correctly that you allow the map $S \to T$ to be non-injective, as long as the composite $R \to T$ is injective?

@Gupta If a power series $F$ has positive convergence radius then also any power series that can be reasonably expressed in terms of $F$ (for example as a rational function of $F$ and its derivatives and integrals and arbitrary compositions of those when they make sense) also will a positive radius of convergence. So if $G$ as zero radius of convergence, then is cannot be reasonably expressed in terms of $F$. And Gerald's example isn't just an isolated counter-example: this behaviour happens whenever $f(n)$ grows exponentially with $n$, so includes many interesting sequences.

Thanks! This argument is quite famous and I have actually seen this argument before, so I should have known that what I was trying to do had no chance of ever working...

Thanks! You are right, the question is subtly but seriously problematic. I actually mentioned in the question already that the action of $F_p$ is well-determined only up conjugation, but somehow I did not realize that this makes the question of whether $F_p(\Lambda_\ell) \subset \Lambda_\ell$ completely meaningless in most cases.

I'm confused, why is $Fr$ invertible?

Just out of curiosity, what is an example of a group that is not algebraically good?

@burtonpeterj This sum is sometimes called the Mertens function. According to Wikipedia, there is a conjecture by Steve Gonek that this sum should by $O(\sqrt{n} (\log \log \log n)^{5/4})$, so it would grow even a little bit slower than the coin flips heuristic would suggest.

Isn't the special fiber of that map given by $\operatorname{Spec} k[x]/(x^2) \to \operatorname{Spec} k$?