a) Let $x$ be an invertible element of $R$, let $e_1$ and $e_2$ be idempotent elements of $R$ such that $x = e_1e_2$, and notice that $1 - e_1 = (1 - e_1) \cdot 1 = (1 - e_1)e_1e_2x^{-1} = (e_1 - e_1^2)e_2x^{-1} = 0$, so $e_1 = 1$ and $x^2 = e_2^2 = e_2 = x$. Since $x$ is invertible, it follows that $x = 1$.

b) Begin by noticing that there are no non-zero nilpotent elements in $R$. Indeed, if $x$ is an element of $R$ and $x^k = 0$ for some integer $k > 1$, then $1 - x^k = (1-x)(1+x+\dots+x^{k-1})$, so $1-x$ is invertible and $x=0$ by (a).

Next, we show that every idempotent element of $R$ is central (it commutes with every element of $R$). Let $e$ be idempotent, let $x$ be any element of $R$ and write
$$(ex - exe)^2 = exex - exexe - exeex + exeexe = exex - exexe - exex + exexe = 0,$$
to deduce that $ex = exe$, by the preceding. Similarly, $xe = exe$, so $ex = exe = xe$.

Finally, let $x$ be an element of $R$, and write $x = e_1e_2$, where $e_1$ and $e_2$ are idempotent. By the preceding, $e_1e_2 = e_2e_1$, so $x^2 = (e_1e_2)^2 = e_1^2e_2^2 = e_1e_2 = x$.