For a normal subgroup $N$ of $G$, we have $aN = N$, then it means $a \in N$, and a similar conclusion holds in a field with multiplication. In both cases, it is due to the existence of an inverse. When working with quotient groups, this seems to be very applicable in that we can conclude $a-b \in N$ from $a+N = b+N$.
However, in a commutative ring $R$ with unity, let $I$ be an ideal in $R$, and $r \in R$, if we have $rI = I$, can we conclude anything?
In this case, we don't have inverses, so I'm a little unsure what to say. When $r$ is a unit, this is of course true, but I'm not sure if $rI =I$ necessarily implies that $r$ is a unit, and whether there is something to say about $r$ in general.
Let $R$ be a commutative ring and $I$ be a nonzero ideal of $R$, and let $r\in R$. As noted by the OP, if $r$ is a unit, then $rI=I$. The converse result is true in some but not all cases.
We'll show $rI=I$ implies $r$ is a unit if (i) $R$ is a UFD or (ii) $I$ is finitely generated and has annihilator zero (e.g., $R$ is an integral domain). To show (i) is true, let $i_0$ be a nonzero element of $I$ with shortest possible factorization. If $r$ is not a unit, then $ri$ has a longer factorization than $i$ for any nonzero $i\in I$. Thus $ri$ can never equal $i_0$ and so $rI\neq I$. To show (ii) is true, we invoke Nakayama's Lemma. This tells us that if $rI=I$ (and $I$ is finitely generated), there is an $s\in R$ with $(1-rs)I=0$. If $I$ has zero annihilator, this implies $r$ is a unit.
More generally, we've shown that if $I$ is finitely generated, then $rI=I$ if and only if there is an $s\in R$ with $(1-rs)I=0$. Such elements $r$ need not be units; if $I$ is not finitely generated, this criterion is no longer valid. Here are a couple of examples. First, let $R$ be the integers mod $6$. If $I=\bar3 R$ and $r=\bar 3$, then $rI=I$, but $r$ is not a unit. If $I=\bar2 R$ and $r=\bar 2$ or $r=\bar 4$, then $rI=I$, but $r$ is not a unit or an idempotent.
Next, consider $R=\mathbb{Z}+x\mathbb{Q}[x]$, which is clearly an integral domain. Let $I=x\mathbb{Q}[x]$: this is an ideal of $R$. If $r$ is any positive integer, $rI=I$, but if $r>1$, $r$ is not a unit.