Binary operation

The groups will be isomorphic, but don't forget it's asking you to count the number of binary operations that form a group, not the groups themselves.

What makes you think you can define two binary operations on a group of order 2? What are they? [There is only one binary operation you can define.]

To define a binary operation you need to draw out a multiplication table. Say your group is $\{e, x \}$, and $*$ is some binary operation. Then you have no choice over what $e*e, e*x, x*e$ are equal to because $(G, *)$ has to satisfy group axioms. But you have two "choices" over what $x*x$ can be: it can either be $e$ or $x$. However, one of these leads to a contradiction, so it turns out you only have one choice over what $x*x$ can be. And hence there is only one binary operation. [Or at least, if $\circ$ is another binary operation then the map $g \mapsto g$ from $(G, *)$ to $(G, \circ)$ is an isomorphism.]

http://answers.yahoo.com/question/index?qid=20091013102041AAVgcah

This link is confusing then to what you have said!

D'oh! I never considered treating $x$ as the identity.

So the link is correct?

Yup. I fell into the trap of naming the elements of the group suggestively; i.e. calling one of them 'e'. Instead what you should do is let G={x,y}, and then for each choice of identity (either e=x or e=y) there is just one binary operation (as in my post), but since you have two choices of identity and there is one binary operation for each, there are two choices of binary operation. The multiplication tables make it fairly obvious whether or not the two are isomorphic.