We don't need that "two"; the result still holds if we assume that every element of the ring is a product of "some" idempotent elements of the ring. Let $R$ be the ring, which we assume has $1$ and suppose that $ab = 1$ for some $a,b \in R.$ We claim that $a = b = 1.$ Suppose, to the contrary, that $a \ne 1$ and let $a = e_1e_2 \cdots e_k,$ where each $e_j$ is an idempotent and $e_1 \ne 1.$ Let $c = e_2 \cdots e_kb.$ Then $e_1c =ab= 1$ and hence $$1 - e_1 = (1 - e_1)e_1c =e_1c-e_1^2c=e_1c-e_1c= 0,$$which gives $e_1 = 1,$ contradiction! So $a=b=1$ and that solves the first part of your problem. Now suppose that $x^2 = 0$ for some $x \in R.$ Then $(1 - x)(1 + x) = 1$ and therefore $x = 0$, by the first part of the problem. Now see that $$(ex-exe)^2=(xe-exe)^2=0$$for any idempotent $e \in R$ and any element $x \in R.$ Thus $ex = xe$ and so $e$ is central. Hence, since every element $x \in R$ is a product of some idempotent elements, we have $x^2=x$ and that solves the second part of your problem.

for the first part, this is another soln. let $u$ be a unit. $u=ab$ , a , b idempotent $a^2=a \implies a=1$ similarly $b=1$ $u=ab=1$ for the second part, $(-1)^2=1 \implies -1=1 \iff 2=0 \iff char(R)=2$ i tried to show that R is commutative by $(xy-yx)^2=0$ but $(xy-yx)^2=xyxy+yxyx-xyx-yxy$ , which is where i got stuck... 2nd part of problem is related to: https://artofproblemsolving.com/community/c7h1060323

a) If $a$ and $b$ are idempotents and $ab$ is a unit, we have \[ a=a(ab)(ab)^{-1}=(ab)(ab)^{-1}=1 \]and similarly $b=1$, so $ab=1$. $\square$ b) Since $-1$ is a unit, we must have $-1=1$ so $2=0$. If $x^2=0$, then $(x+1)^2=x^2+1=1$ implies that $x+1$ is a unit, so $x=0$. Thus for any idempotents $a$ and $b$, we have \[ (ab+aba)^2=abab+ababa+aba^2b+aba^2ba=0 \]so $ab=aba$. Hence $abab=ab^2=ab$, as desired. $\square$