Division ring if and only if it has no proper left ideals.

$(rz)r=r(zr)=r$, thus $r(1-zr)=0$, $zr=1$.

Remark that the fact that $zr=1$ follow from the fact that $R$ does not have divisors of zero: if $ab=0$ $a,b\neq 0$ , since $Ra=R$, $ra=1$, $r(ab)=(ra)b=b=0$. Contradiction.

You have the side switched, but it's a minor detail.

If $r\ne0$, then $Rr=R$, so you find $z$ with $zr=1$. Next, also $Rz=R$, so there is $s$ with $sz=1$.

Now prove $r=s$.