It seems like the proof that $(\mathfrak{a}:\mathfrak{b}) = \{ x \in R: x b \in \mathfrak{a} \ \ \forall b \in \mathfrak{b} \}$ is an ideal does not actually depend on $\mathfrak{b}$ being an ideal, and instead only requires $\mathfrak{a}$ to be an ideal.
Question: Is this a correct conclusion? I.e. is the quantity $(\mathfrak{a}:S)$ well-defined for any ideal $\mathfrak{a}$ and any subset $S$ of a commutative ring with unit $R$? $\newcommand{\ideal}{(\mathfrak{a}:\mathfrak{b})}$
Attempt:
(i) $(\mathfrak{a}:\mathfrak{b})$ is an abelian subgroup: given $x, y \in \ideal$, we have for any $b \in \mathfrak{b}$ that $\mathfrak{a}$ an ideal means that $xb, yb \in \mathfrak{a} \implies xb - yb = (x-y)b \in \mathfrak{a}$. Therefore, $x,y \in \ideal \implies x-y \in \ideal$, using only the fact that $\mathfrak{a}$ is an ideal.
(ii) $\ideal$ is absorptive: we have for any $b \in \mathfrak{b}$ that, since $\mathfrak{a}$ is an ideal, $xb \in \mathfrak{a} \implies r(xb)=(rx)b \in \mathfrak{a}$ for all $r \in R$. Therefore $x \in \ideal \implies rx \in \ideal$ for all $r \in R$, and thus $\ideal$ is an ideal.
Since $\ideal$ is an abelian subgroup and absorptive, it is an ideal. In the above, we only appealed to the fact that $\mathfrak{a}$ is an ideal, and whether $\mathfrak{b}$ is an ideal or not does not seem to matter to the conclusion that $\ideal$ is an ideal. In particular, for any ideal $\mathfrak{a}$, it seems that $(\mathfrak{a}:\{1\}) = \mathfrak{a}$.
If $\frak{a}$ is a left ideal and you want $(\frak{a}:\frak{b})$ to be a left ideal, or if you are working in a commutative ring, this is fine.
In a non-commutative ring, if you want $(\frak{a}:\frak{b})$ to be a two-sided ideal, you need $\frak{a}$ and $\frak{b}$ to be left ideals so that you can handle both $rx$ and $xr$ when $x\in(\frak{a}:\frak{b})$. (Some weaker conditions than left ideal work for $\frak{b}$, but you do need a condition on $\frak{b}$ in the non-commutative case if you want the result to be a two-sided ideal.)