jmc

From the addition problem, we know that in the units digit, it's impossible to have $B+A=0_7$, so it must be that $B+A=10_7=7$. That means we carry a 1 to the next column and get $A+B+1=A{A}_7$. Since $B+A=10_7$, then $A+B+1=11_7$ and $A$ represents the digit 1. That tells us that $B+1=7$, so $B$ represents the digit 6. The product of $A$ and $B$ is $\boxed{6}$.