As pointed out by Makoto, on this question about power series rings and the axiom of choice, an idea I had needed the axiom of dependent choice to work. However, the construction raises another interesting problem which I'm hopeful is easier to solve.

Let $A$ be a commutative ring with $1$. Let $I\in A[[x]]$ be a finitely generated ideal of the power series ring. If we don't assume the axiom of dependent choice, is it possible to have an element $f(x)\in A[[x]]\setminus I$ such that for every $n\in \mathbb{N}$ there exists $f_n(x)\in I$ with $x^n|(f(x)-f_n(x))$?

Intuitively, this says that we can build a linear combination, from the generators of $I$, which matches $f$ to any *finite* degree, but there is no combination that matches $f$ completely. If we write $I=(g_1(x),\ldots, g_k(x))$ and we let $I'$ be the ideal generated by the coefficients of all the $g$'s and $f$, I'm most interested in the case when the leading terms of $g_1(x),\ldots, g_k(x)$ generate $I'$. In this case, the linear combinations can be built up in compatible ways (degree by degree), but could it still not be possible to get all of $f$?

Edited to add: Here is an even simpler question. Let $a\in A$ and let $f(x)\in A[[x]]$ be such that every coefficient of $f(x)$ lives in the ideal generated by $a$. Can we say that $f$ lives in the ideal generated by $a$, without using a choice axiom?