Trick to proving a group has exactly one idempotent element - Fraleigh p. 48 4.31 

If $*$ is a binary operation on a set $S$, an element $x \in S$ is an idempotent for $*$ if $x * x = x$.

Let $\langle G, *\rangle$ be a group and let $x\in G$ such that $x*x = x.$ Then $x*x = x*e$, and by left cancellation,  $x = e$, so $e$ is the only idempotent element in a group.

The trick here looks like writing $x$ as $x*e$. How can you prognosticate (please see profile) this? I didn't see it. It also looks like you have to prognosticate the 'one idempotent element' to be the identity element. Is this right? Can someone make this less magical and psychic? 

This does not look so much like prognostication to me. The hint is trying to make explicitly clear something that is happening. You could instead start by right multiplying both sides by $x^{-1}$. Then you get $xxx^{-1} = xx^{-1} \implies x(e) = e \implies x = e$. By the uniqueness of the identity, there is only one such element. 
