Suppose $aR=R$ for all nonzero $a\in R$, and $R$ is unital. Fix $a\in R$; we seek an inverse. By hypothesis we know $1\in aR$ so $1=ab$ for some $b\in R$. And similarly $bR=R$ so $1=bc$ for some element $c\in R$. But then we have $a=abc=c$, so $1=ab$ and $1=ba$ and so arbitrary $a\in R$ have two-sided inverses, and hence $R$ is a division ring.
A symmetrical argument holds if we suppose $R=Ra$ for all nonzero $a\in R$.
A unital ring $R$ may fail to have two-sided ideals besides $0$ and $R$ and still fail to be a division ring due to the presence of zero divisors. The textbook examples are the matrix rings $M_n(k)$ (for any integer $n>1$) over any scalar field $k$ (see this question). Indeed if $k$ is finite then so is $M_n(k)$!