@LSpice: we are assuming that $R^\times=\{1,u,\ldots,u^4\}$, so we must have $-u=u^i$ for some $i$. Since $u$ is a unit, this means that $-1=u^{i-1}$ for some $i$. But $-1$ has order dividing $2$, and we are in a cyclic group of order $5$, so $-1$ must have order $1$, i.e. be equal to $1$. I guess, I could have bypassed the whole $-u$ business. Let me edit.

@MC: You are welcome!

What is the "usual inner product"?

This reasoning is not complete, asymptotics do not tell you everything. A counterexample is $S=T=\{\text{even integers}\}$.

There is no "best result in that direction", since you have too few hypotheses. For example if $K/\mathbb{Q}$ is finite and the representation is over a field of characteristic $0$, then everything is diagonalisable, but that has nothing to do with Galois representations. In general, (Weil-Deligne) representations for which Frobenii are diagonalisable are called "Frobenius semisimple". You can google that term for lots of literature, but almost none of it will be "elementary".

You have almost no hypotheses, so the answer is clearly no. Just take $K/\mathbb{Q}$ to be finite Galois, e.g. with Galois group $S_3$, and take $\rho$ to be irreducible over a finite field such that some element of $S_3$ is sent to a non-diagonalisable matrix. For $S_3$, the standard representation arising from the isomorphism $S_3\cong {\rm GL}_2(\mathbb{F}_2)$ will do, where the involutions are not diagonalisable. By Chebotarev, every element of your Galois group will be a Frobenius at some $p$.