I don't know if this has been solved in the literature, but here is an example showing that the condition is not left-right symmetric.
Let $$ R=\mathbb{F}_2\langle a,b,c\, :\, a^2=ab=ba=b^2=0,\ ca=b,\ cb=a\rangle. $$
The ideal generated by $a$ and $b$ is nilpotent of index $2$; call that ideal $J=(a,b)$. We see that $R/J\cong \mathbb{F}_2[c]$ is an integral domain and Jacobson semisimple, so $J$ is the Jacobson radical of $R$. As the only unit in $\mathbb{F}_2[c]$ is $1$, and an element of $R$ is a unit if and only if its image is a unit modulo the Jacobson radical, we see that $U(R)=1+J$.
Note that $Ra=Rb$ (since $ca=b$ and $cb=a$). However, $U(R)a=a\neq b$, so $R$ is not a left UGP ring.
Now, suppose that $rR=sR$ for some elements $r,s\in R$. (We can quickly reduce to the case that they are nonzero.) The relations allow us to write $r$ uniquely in the form $r=f(c)+ag(c)+bh(c)$ for unique polynomials $f(c),g(c),h(c)\in \mathbb{F}_2[c]$. Similarly, $s=f'(c)+ag'(c)+bh'(c)$. (These prime marks are just to distinguish polynomials, and have no reference to derivatives.)
Now, the elements of $rR+J$ are exactly $f(c)\mathbb{F}_2[c]+J$, and similarly the elements of $sR+J$ are exactly $f'(c)\mathbb{F}_2[c]+J$. This means $f(c)\mathbb{F}_2[c]=f'(c)\mathbb{F}_2[c]$. As $\mathbb{F}_2[c]$ is a UGP ring, and the only unit is $1$, we have $f(c)=f'(c)$.
There are two main cases: First, suppose that $f(c)=f'(c)=0$. Then, since $aJ=bJ=0$, we see that $$ rR=(ag(c)+bh(c))(\mathbb{F}_2[c]+J) = (ag(c)+bh(c))\mathbb{F}_2[c]=ag(c)\mathbb{F}_2[c] + bh(c)\mathbb{F}_2[c]. $$ A similar fact holds for $sR$, and so an argument similar to the one in the previous paragraph implies that $g(c)=g'(c)$ and $h(c)=h'(c)$. Hence $r=s$.
Finally, suppose that $f(c)=f'(c)$ is nonzero. From $rR=sR$, we know that $r=st$ for some $t\in R$. Modulo $J$, we must have $t=1$. But then $t$ is a unit.