Here's my attempt :
The unity of $A$ and the unity of $B$ are both same, so, we'll call it $1$ unambiguously.
$( \Leftarrow )$ Suppose $A$ has characteristic $p$ then $p \cdot 1 = 0$. Since $1$ is also in $B$, it must be that characteristic of $B$, say, $p_{B}$ divides $p$. Also, $p_{B} \ne 1$ thus $p_{B}=p$.
$( \Rightarrow )$ Suppose $B$ has characteristic $p$ then $p \cdot 1 = 0$. Since $1$ is also in $A$, it must be that characteristic of $A$, say, $p_{A}$ divides $p$. Also, $p_{A} \ne 1$ thus $p_{A}=p$.
Is the proof correct? This problem is in Pinter's A book of Abstract Algebra.