Let $R$ be a commutative ring with unit.
On the elements of $R$ we can define the following relations:$$a\sim_1b\iff a=ub\text{ for some unit }u$$
and:$$a\sim_2 b\iff r,s\in R\text{ exist with }a=rb\text{ and }b=sa$$
It is clear to me that both relations are equivalence relations and that $a\sim_1b\implies a\sim_2b$.
Also it is clear to me that the opposite of this is true if $a$ or $b$ is not a zero-divisor.
Questions:
1) Do we always have $a\sim_2b\implies a\sim_1b$? (I suspect not)
2) If not then can you provide a counterexample?
3) Is the condition "$R$ has no zero-divisors" (sufficient for $a\sim_2b\implies a\sim_1b$) also a necessary condition for $a\sim_2b\implies a\sim_1b$?
Thank you in advance and sorry if this question is a duplicate.
As you suspect, the answer to your first question is no. A counterexample is $R=\mathbb{Z}[X,Y,Z]/(X(1-YZ))$.
One may check that the units of $R$ are $\pm 1_R$. I let you find two elements generating the same ideal but which do not differ by a unit.
For your last question, the answer is NO as well. The ring $\mathbb{Z}/4\mathbb{Z}$ has zero divisors, but two elements generating the same ideal necessarily differ by a unit.