Proof Strategies on P versus BPP
Although there aren't any problems known to be $\mathsf{BPP}$-complete (and Sipser gave an oracle relative to which $\mathsf{BPP}$ doesn't have complete problems), one topic to look at here is pseudorandom generators. The existence of a good enough pseudorandom generator implies $\mathsf{BPP} = \mathsf{P}$. This isn't $\mathsf{BPP}$-complete, but it does have the feature that a single algorithm could show $\mathsf{BPP}=\mathsf{P}$.

Also, although not $\mathsf{BPP}$-complete, Polynomial Identity Testing is a good single problem in $\mathsf{BPP}$ to focus on derandomizing. Many problems in $\mathsf{BPP}$ are actually in $\mathsf{RP}$ or $\mathsf{coRP}$, often due to the Schwarz-Zippel Lemma (in fact, there are apparently very few known candidates that are known to be in $\mathsf{BPP}$ but not in $\mathsf{RP} \cup \mathsf{coRP}$, see this related question), so derandomizing PIT would derandomize a whole bunch of problems. Furthermore, derandomizing PIT is known to imply strong lower bounds on $\mathsf{NEXP}$.