Lam informed me that, as far as he knew, this problem was still open. However, the example below shows that the condition is not left-right symmetric.
Let $$ R=\mathbb{F}_2\langle a,b,c\, :\, a^2=ab=ac=ba=b^2=bc=0,\ ca=b,\ cb=a\rangle. $$
The ideal generated by $a$ and $b$ is nilpotent of index $2$; call that ideal $J=\mathbb{F}_2a+\mathbb{F}_2b$. 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, $U(R)a=(1+J)a=a\neq b$, 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 $r=f(c)+\alpha a+\beta b$ for unique polynomial $f(x)\in \mathbb{F}_2[x]$ and unique $\alpha,\beta\in \mathbb{F}_2$. Similarly, $s=g(c)+\gamma a+\delta b$.
Now, looking at $rR=sR$ modulo $J$, we get $f(c)\mathbb{F}_2[c]=g(c)\mathbb{F}_2[c]$. As $\mathbb{F}_2[c]$ is an integral domain, it has the UGP property. Its only unit is $1$, and so $f(c)=g(c)$.
Write $r=st$ for some $t\in R$. If $f(c)\neq 0$, then $t\in 1+J$, and hence $t$ is a unit. If $f(c)=0$, then $r,s\in J$, and so $$ r\mathbb{F}_2 =rR=sR=s\mathbb{F}_2. $$ Thus $r=s$ in this case.