We start off by replacing $ C $ with $ A+B $ and changing the form of the two-digit integer in the second equation. \begin{align*} 10A+A-B&=2\times(A+B)\quad\Rightarrow\\ 11A-B&=2A+2B\quad\Rightarrow\\ 9A&=3B\quad\Rightarrow\\ 3A&=B \end{align*}Now we replace $ C $, change the two-digit integer, and then substitute $ B $ with $ 3A $ in the third equation. \begin{align*} (A+B)\times B&=10A+A+A\quad\Rightarrow\\ &=12A\quad\Rightarrow\\ (A+3A)\times3A&=12A\quad\Rightarrow\\ (4A)\times3A&=12A\quad\Rightarrow\\ 12(A)^2&=12A \end{align*}For $ (A)^2 $ to equal $ A $, $ A $ must equal 1. Since $ 3A=B $, $ B=3 $. That means $ A+B=C=4 $. So the sum of the three digits is $ 1+3+4=\boxed{8} $.