Suppose $R$ is a commutative ring without identity, and let $a, b \in R$. Is it true that $b \in (a) \iff b=ar$ for some $r \in R$?
This seems to be implied (I can provide the details upon request) in Dummit & Foote.
The definition of $(a)$ provided in Dummit & Foote is the intersection of all ideals which contain $a$. In commutative rings with identity, it is proved that $(a) = R \{a \} R = \{r_1ar_2 + \cdots + \cdots r_nar_n| \text{ where each } r_i \in R \} = \{ar | r \in R \}.$ In this case, it's easy to prove that $b \in (a) \iff b=ar$ for some $r \in R$.
However, I think this is not true in rings without identity, and I think I have a counterexample:
Let $R = 2 \mathbb{Z}$, and $a=6$. The only ideals of $R$ which contain $6$ are $2\mathbb{Z} = R$ and $6\mathbb{Z}$, so the intersection of all ideals which contain $6$ is $6\mathbb{Z}$.
Now $18 \in (6) = 6\mathbb{Z}$, but $18 \not = 6r$ for some $r \in 2\mathbb{Z}$.
EDIT: This is where I found the statement in D & F. I have the third edition, paperback. On page $274$, we have the definition of divisibility:
Definition: Let $R$ be a commutative ring and let $a,b \in R$ with $b \not = 0$.
$\vdots$
Note that $b|a$ in a ring $R$ if and only if $a \in (b)$ ...
(Sorry I reversed the roles of $a, b$ in my question)
Edit: Hat tip to Carl Mummert, who pointed out that my original answer used the wrong definition of the ideal generated by $a \in R$.
According to Dummit and Foote, the ideal generated by an element $a$ of a ring $R$, denoted $(a)$ is the smallest ideal containing $a$. When $R$ is a commutative ring with identity, then $(a) = Ra$, where $Ra = \{r a : r \in R \}$ is also an ideal.
The example you provide shows that $(a) \neq Ra$ in general when $R$ is a commutative ring without identity. Indeed, $(6)$ consists of all multiples of $6$, while $R6$ consists of all multiples of $12$ in your example.
You are correct that Dummit and Foote's statement is wrong in general. The correct version is $b \mid a \Leftrightarrow a \in Rb$, for any commutative ring $R$. When $R$ contains an identity, this is equivalent to $b \mid a \Leftrightarrow a \in (b)$, but their statement is strictly speaking false without the assumption that $R$ contains an identity, and your counterexample shows that. In your example, $18 \in (6)$ but $6 \nmid 18$ in $R = 2\mathbb{Z}$.