Showing that $1$ is the only unit in a ring with identity $R$ such that $a^2 = a$ for all a in $R$ So I've been given a ring $R$ with identity (no further information on what the identity is or what kinds of elements are in $R$ or operation definitions) such that $a^2 = a$ for all $a \in R$. I need to show that the only unit in $R$ is $1$.
I've kind of assumed a proof by contradiction approach. I'm assuming there are other units in $R$ which I've called $u$, called its inverse $x$ and called the identity $I$. So I've done the following
$ux=1$
$(ux)I=I$
$ux=I$
$u^2=u$
$(u^2)I=uI$
$u^2=u(ux)$
$u^2=(u^2)x$
$1=x$
$ux = 1$
$u(1) = 1$
$u = 1$
Contradiction, $1$ is the only unit in $R$.
I know I've got the right answer but I'm a little apprehensive about it. Do I need to do the left-side multiplications too, since there's nothing about commutativity in the original question?

You have all the right pieces. It's just overly complicated. Let $1$ be the multiplicative identity of $R$ and assume $u$ is some unit of $R$. Then
$$u = u1 = u(uu^{-1}) = u^2u^{-1} = uu^{-1} = 1,$$ so $u = 1$ is the only unit in $R$.
