Turning my comments into an answer:
First off, it's important to state the incompleteness theorem carefully (to be precise, I'm giving Rosser's improvement of Gödel's first incompleteness theorem):
There is no first-order theory $T$ which is computably axiomatizable, consistent, complete, and interprets Robinson arithmetic.
Of particular note are the phrases "computably axiomatizable" and "interprets Robinson arithmetic." These are technical terms which I won't define here, but I will very briefly motivate them:
Computable axiomatizability is basically a "simplicity" condition, demanding that the theory in question not be so complicated that we can't even write down a set of axioms for it in a reasonable way.
The interpretability criterion is a "strength" condition. Our theory needs to be able to implement basic arithmetical calculations.
Basically, both too complicated theories (such as the set of all true sentences about $(\mathbb{N};+,\times)$) and too weak theories (such as Presburger arithmetic) are beyond the scope of applicability of the incompleteness theorem.
This isn't really related to your specific question, but I think being careful about what the theorem actually says will demystify it substantially.
OK, now on to the actual question.
There are two ways a theorem (including Gödel's) can be wrong: either the argument used is faulty, or the axioms used are not all correct. Here for simplicity I'm adopting a "naive Platonist" position, that there is a "true mathematics" out there even if we don't know what it is.
There isn't much to say about the first possibility; the incompleteness theorem's proof has been checked ad nauseam, including implemented on computers, and additional proofs have been discovered as well (see e.g. here). The interesting issue is the second possibility. Because of the rhetoric around the incompleteness theorem, it's quite reasonable to assume that some rather subtle axioms must be at work in its proof (note that I'm speaking of the axioms used in the proof of the incompleteness theorem itself, not the axioms of a system it applies to). This, however, is false: in fact extremely little is needed to run the proof! In order to doubt the incompleteness theorem, we have to be skeptical of rather basic principles of arithmetic itself - specifically, a very weak form of proof by induction. The logical structure of the proof of the incompleteness theorem is quite subtle, but the basic assumptions that argument begins with are just simple properties of arithmetic which we use all the time.
Your understanding is correct, apart from the caveat that there are a lot of details you glossed over. Most significantly, in the part about the second incompleteness theorem, while the first incompleteness theorem shows, 'if S is consistent, then G is true', what we really need is to show that S can prove this... and this is a major undertaking, even on top of the detailed work to show the first incompleteness theorem, and it requires new technical assumptions about S, more detailed than the conditions needed for the first theorem.
But let's put that aside, and address your question. Yes, it certainly feels odd that it is consistent to add "S is inconsistent" as an axiom (sometimes I think that this should be a "paradox" with a name). The reason it feels odd is that you are adding an axiom that is false.
The crucial thing to grasp here is that a theory doesn't need to be correct (according to some standard interpretation) in order to be consistent. In terms of interpretations, the standard interpretation (i.e. the natural numbers) is no longer a model of "S + S is inconsistent", but some of the nonstandard interpretations of S are still models. These nonstandard models have nonstandard natural numbers (larger than any number their initial segment that looks like $\mathbb N$) that they think encode proofs of "0=1" in S.
In "S + S is inconsistent", we also have an odd syntactic situation where for each natural number $n,$ it is provable that "$\mathbf n$ does not code a proof of 0=1 in S" and yet it is also provable that "there is a proof of 0=1 in S". This is called $\omega$-inconsistency. Of course, as in the previous paragraph, we can see that the resolution is that not every element of the domain of an interpretation needs to be represented by a numeral $\mathbf n.$
Perhaps we would like it if it were inconsistent to be wrong, but this would require that our theory decide every sentence (correctly), since, as you've reasoned, if a sentence isn't decidable, you can add either it or its negation and get a consistent system, and both can't be right. And unfortunately this is exactly what Godel's theorem rules out (for sufficiently strong, recursively axiomatizable theories).
Best Answer
First, there is nothing unusual about having a theory with infinitely many axioms. This is the case for both of the two most commonly studied "foundational" theories, Peano Arithmetic and Zermelo-Fraenkel set theory. In both of these cases, what is often described as one named axiom (induction for PA, selection and replacement in ZF) are actually axiom schemas -- that is, each is just a recipe for generating an infinity of different axioms by plugging different logical formulas into the schema.
First-and-a-half, remember that the Gödel procedure for producing an unprovable (and un-dis-provable) statement requires not only that the axioms is effectively generated, but also that the theory is rich enough to be "interesting". (For example, the theory of integers and addition where multiplication is not mentioned is not rich enough for Gödel's construction to work on it).
Second, if your original theory $A_0$ is effectively generated, then your $A_\infty$ will be too -- there's a mechanical, deterministic process that will eventually print every axiom in $A_\infty$.
$A_\infty$ will be consistent too, if $A_0$ is -- by the "compactness" property of formal logic which says that every inconsistent system of axioms has a finite subsystem that's also inconsistent. (If there's a proof of a contradiction in the system, then because proofs are finite by definition, the proof depends on only finitely many of the axioms). Every finite subset of $A_\infty$ will be a subset of one of the $A_n$'s, and all of these are consistent by construction.
Since $A_\infty$ is effectively generated and (because it extends $A_0$) sufficiently rich, we can repeat the Gödel process on it, and get at Gödel sentence for $A_\infty$. Add that to $A_\infty$ to get $A_{\infty+1}$, and proceed ad nauseam. (In this context it is traditional to write $\omega$ instead of $\infty$, and there's a theory of the necessary numbers "beyond infinity" under the name "ordinal numbers").