Question: Let $R$ be a commutative ring, $A$ an ideal of $R$, and $R/A$ an integral domain. Then, as I can see in my book, $(a+A)(b+A)=ab+A=A$. But how does this multiplication exactly go? If $(a+A)(b+A)=ab + aA + Ab + A$ then where does the sum of the terms $aA+Ab$ go?

Also, why does $(a+A)(b+A)=A$ imply that $(a+A)=A$ or $(b+A)=A$?

Thank you for your clarifications.

Answer: 
The multiplication in $R/A$ is *defined* as $(a + A)(b + A) = ab + A$; there is no expansion of brackets.

As for your second question, note that $A$ is the zero element of the ring $R/A$. Now $R/A$ is an integral domain which means it has no zero divisors. Therefore, if $(a + A)(b + A) = A$, then $a + A = A$ or $b + A = A$.
As for where $aA+Ab$ goes: Recall that one of the defining properties of an ideal $A$ is that, for all $x{\in}R$, $xA{\subseteq}A$ and $Ax{\subseteq}A$. Also recall that the definition of $x+A$ is the set {$x+y | y{\in}A$}. So the product ($a+A$)($b+A$) is the SET of all products: {($a+x$)($b+y$)|$x,y{\in}A$}. Expanding each term yields $ab+ay+xb+xy$, where *each of the last three terms* is in $A$, hence so is their sum. I.e, the terms $aA+Ab+AA$ all collapse to their common union, which is only $A$.