Yes. It follows from the Cantor case. First, it is no restriction to assume that $A,B$ are compact totally disconnected (e.g., countable) subsets of the plane and we want to prove that any homeomorphism $A\to B$ extends to a self-homeomorphism of the plane.

Lemma: *every compact totally disconnected subset $K $ of $\mathbf{R}^k$ is contained in a Cantor subset of $\mathbf{R}^k$.*

Proof: we can suppose $K$ nonempty. If $x$ is an isolated point in $K$, let $N_x$ be the distance of $x$ to its complement in $K$. Then choose a Cantor subset $L_x$ in $\mathbf{R}^k$ containing $x$, of diameter $<N_x/2$. Then $L=\bigcup L_x$ is nonempty compact, totally disconnected, contains $K$, and has no isolated point. (Totally disconnected requires a little argument: the point is to check that for every partition of $K$ into 2 clopen subsets $U,V$, there exists only finitely many $x$ such $L_x$ meets both $U$ and $V$, using that $N_x$ tends to zero and $U,V$ are at positive distance. Using this, it's easy to extend continuously every continuous map $K\to\{0,1\}$ to $L$). So it's a Cantor set. $\Box$

Since it is known that every Cantor subset of the plane can be mapped by a self-homeomorphism of the plane onto the standard dyadic Cantor set in the line $\mathbf{R}\times\{0\}$ (This follows from the Denjoy-Riesz theorem, see this MathSE post), we can suppose $A$, $B$ are contained in this given dyadic Cantor set.

So now consider $A,B$ in $\mathbf{R}\times\{0\}$. Then if $Q$ is a Cantor set, clearly every homeomorphism $A\to B$ extends to a homeomorphism $A\times Q\to B\times Q$. Assuming $A$ nonempty, $A\times Q$ is a Cantor set. Hence if we have proved the extension result in the Cantor case, we are done.

So after this second reduction, we can assume $A=B$ is the standard dyadic Cantor set in $\mathbf{R}\times\{0\}$. This case is not obvious, but is already answered in this other MathSE post, with a reference to Edwin E. Moise's book *Geometric Topology*.