The axiom of choice is in fact equivalent to the assertion that every commutative unital ring has a maximal ideal.
Since the negation of the axiom of choice is as non-constructive as the axiom of choice itself, we can only say that there exists a commutative unital ring without a maximal ideal when the axiom of choice fails. The sets which are involved in this process are non well-orderable.
It is important to understand that much like the axiom of choice only assures that certain objects exist; its negation has a very similar intangible action: it only assures that some sets cannot be well-ordered, some partial orders in which every chain is bounded will not have maximal elements, and some families of non-empty sets do not have a choice function. We have no means of knowing where these objects are without further assumptions.
It is consistent that the axiom of choice holds for every set you have ever dreamed using, but then fails acutely. In that universe you cannot imagine to actually find that set which cannot be well-ordered, or that ring without a maximal ideal, and so on. But it is also consistent that the axiom of choice fails "nearby" and the counterexamples appear in relatively familiar sets (objects related to, or defined from the real numbers for example).
The best way to actually "find" a ring without a maximal ideal is to follow the proof of how the existence of maximal ideals imply the axiom of choice. These proofs often consists of taking a family of non-empty sets, defining some ring (or whatever) and using the maximal ideal to prove the existence of a choice function. Therefore starting with a family of non-empty sets which does not have a choice function guarantees that the process fails and that the ring defined in such proof will not have any maximal ideals.
Added:
Some point that came up in my comment exchange with Trevor Wilson under his answer, is that the axioms are syntactical. They allow us to write proofs. It seems to me, upon re-reading this question that you think something like:
Take a universe of ZFC, $R$ is a unital ring and $I$ is a maximal ideal. Now just don't assume that AC holds. Now we can't prove that $I$ is a maximal ideal.
That's false for two main reasons:
Once you fixed a universe of sets, the axiom of choice is either true or false in that universe. Even if you don't assume it, it has a truth value, and since we began from a universe of ZFC this truth value is indeed true. We just might not be able to write a proof from ZF that $I$ is maximal, but this is something that is still true in that universe of sets.
So changing the assumptions does not necessarily mean that you have changed your universe of sets.
If you have a definition for a ring (e.g. $\Bbb{R^R}$ with pointwise addition and multiplication) then the actual underlying set, and more importantly its subsets, may change between one universe of set theory and another. So for example we cannot prove from ZF that $\Bbb{R^R}$ has a maximal ideal because there are universes of set theory where this is false. But the underlying set $\Bbb{R^R}$ is very different between those universes, and again more importantly, its power set is different.
So to sum up all my answer (with its two additions), simply removing the assumption that the axiom of choice holds will not falsify the axiom of choice. It might be that you just won't be able to write a proof that there exists a maximal ideal in every unital ring.
If, however, you allow yourself to change the universe of set theory then it is possible that a certain ring will "lose" its maximal ideals, simply because we removed those sets from the universe (and in some cases did a whole revamp of the universe altogether).
You are working only with finite products, and this hold in general. In the Choice Function $\Rightarrow$ AoC direction you don't know if $D$ is nonempty, that is what you are trying to pare working only with finite products, and this hold in general. In the AoC $\Rightarrow$ Choce Function direction you need a choice function on $\mathcal{A},$ not in $D.$ Moreover, I think we need to clarify some definitions.
First one definition
Definition: Let $\mathcal{A}$ be a nonempty collection os sets. A surjective function $f$ from some set $J$ to $\mathcal{A}$ is called an indexing function for $\mathcal{A}.$ $J$ is called the index set. The collection $\mathcal{A},$ together with $f,$ is called an indexed family of sets.
I'll use your definition of axiom of choice.
Axiom of choice: For any indexed family $\{ A_\alpha \}_{\alpha \in J}$ of nonempty sets, with $J \neq 0,$ the cartesian product
$$ \prod_{\alpha \in J} A_\alpha$$
is not empty. Recall that the cartesian product is the set of all functions
$$ \mathbf{x}:J \to \bigcup_{\alpha \in J}A_\alpha$$
such that $\mathbf{x}(\alpha) \in A_\alpha$ for all $\alpha \in J.$
Now,
Existence of a choice function: Given a collection $\mathcal{A}$ of nonempty sets, there exists a function
$$ c: \mathcal{A} \to \bigcup_{A \in \mathcal{A}}A$$
such that $c(A)$ is an element of $A: c(A)\in A$ for each $A \in \mathcal{A}.$
Axiom of choice $\Rightarrow$ Existence of choice function.
We are assuming that for any indexed family $\{ A_\alpha \}_{\alpha \in J}$ of nonempty sets, with $J \neq 0,$ the cartesian product $ \prod_{\alpha \in J} A_\alpha$ is not empty. Let $\mathcal{A}$ be a collection of nonempty sets. We have to prove that there exists a function $ c: \mathcal{A} \to \bigcup_{A \in \mathcal{A}}A$ such that $c(A) \in A$ for each $A \in \mathcal{A}.$ We first index $\mathcal{A}$: let $J=\mathcal{A}$ and $f:\mathcal{A} \to \mathcal{A}$ given by $f(A)=A.$ Then $\{A\}_{A \in \mathcal{A}}$ is an indexed family of sets, so we can consider its cartesian product $\prod_{A \in \mathcal{A}}A.$ By hypothesis, this product is nonempty, so there exists a function
$$ \mathbf{x}: \mathcal{A} \to \bigcup_{A \in \mathcal{A}}A$$
such that $\mathbf{x}(A) \in A$ for each $A \in \mathcal{A}.$ Then $c=\mathbf{x}$ is the function we were looking for.
Existence of choice function $\Rightarrow$ Axiom of choice
Now we are assuming that the existence of choice function, and let $\{A_\alpha \}_{\alpha \in J}$ be an indexed family of nonempty sets, with $J \neq 0.$ This means that there is a nonempty collection of sets $\mathcal{A}$ and there is an indexing (i.e. surjective) function $f:J \to \mathcal{A}$ such that $f(\alpha)=A_\alpha \in \mathcal{A}$ for each $\alpha \in J.$ By the existence of choice function, there is a function $c:\mathcal{A} \to \bigcup_{A \in \mathcal{A}}A$ such that $c(A) \in A$ for each $A \in \mathcal{A}.$ Thus
the function
$$ \mathbf{x}:= c \circ f : J \to \mathcal{A} = \bigcup_{\alpha \in J}A_\alpha$$
satisfies $\mathbf{x}(\alpha)=c(f(\alpha))=c(A_\alpha) \in A_\alpha$ for each $\alpha \in J,$ so the product $\prod_{\alpha \in J}A_\alpha$ is nonempty.
Best Answer
This is a very delicate point about the context of domain and codomain, which in set theory exist as an external properties we give functions, rather than internal properties of them (as in category theory).
This means that we first fix some domain and codomain and we can talk about functions from $A$ to $B$. Now we can say that $f$ is surjective if its range is all $B$. Note that without this context there is no codomain, and every function is onto its range, therefore the term "surjective" is meaningless.
Now when we say that a function $f$ has a right inverse we mean that there is $g\colon B\to A$ such that $f\circ g=\text{id}_B$. When we say that $f$ has a left inverse we mean there is such $g\colon B\to A$ for which $g\circ f=\text{id}_A$. Without the axiom of choice we can prove that if a function has a right inverse then it is surjective, and that $f$ is injective if and only if it has a left inverse.
The axiom of choice is equivalent to saying that whenever $A$ and $B$ are two sets, the $f$ is a surjective function if and only if it has a right inverse. But we had set the context first. Of course we can omit it, because we are talking about any set, so we may as well replace $B$ by the range of $f$. But some context must be set for otherwise the term "surjective" is meaningless.
On the other hand, the very basic theorem that $f\colon A\to B$ has an inverse if and only if it is a bijection has nothing to do with the axiom of choice. The statement is really saying that $f$ is a bijection if and only if there is some $g\colon B\to A$ which is both right and left inverse of $f$. From the above remarks you can see that without the axiom of choice if there is an inverse (both left and right, that is) then the function is a bijection; but if it is a bijection then we can construct its left inverse and show that it is also a right inverse.