Are omega-consistent extensions of PA always consistent with each other? The question is as in the title. In the edit history you can find my attempt to formalise the question, but that was a failure, for reasons stated clearly in the comments. Thus, my question is just:
Are omega-consistent extensions of PA always consistent with each other?
 A: $\def\pa{\mathrm{PA}}\def\N{\mathbb N}\DeclareMathOperator\Th{Th}\def\pri{\mathrm{Pr}_1}\def\code#1{\ulcorner#1\urcorner}$The answer is no: in fact, there are sentences $A$ such that $\pa+A$ and $\pa+\neg A$ are both $\omega$-consistent, while together they are visibly inconsistent.
One way to see this is to use the bounded arithmetic complexity of $\omega$-consistency: the set
$$\{A: \text{$\pa+A$ is $\omega$-inconsistent}\}$$
is $\Sigma_3$ (“there exists a formula $B(x)$ such that for all $n\in\N$ there exists a PA-proof of $A\to B(n)\land\exists x\,\neg B(x)$”), hence the set
$$S=\{A: \text{$\pa+A$ is $\omega$-consistent}\}$$
is $\Pi_3$. Also, if $\N\models A$, then $\pa+A$ is $\omega$-consistent; that is, $S\supseteq\Th(\N)$. However, these two sets cannot be equal, as $\Th(\N)$ is not $\Pi_3$ (in fact, it is not arithmetically definable at all). Thus, $S\supsetneq\Th(\N)$, i.e., there exists a sentence $A$ such that $\pa+A$ is $\omega$-consistent, but $\N\models\neg A$. The latter means that $\pa+\neg A$ is also $\omega$-consistent, QED. The argument actually shows that one can find a $\Sigma_3$-sentence $A$ with this property. (This is best possible: if $A$ is $\Sigma_2$, only one of $\pa+A$ and $\pa+\neg A$ is $\omega$-consistent.)
Another way to prove this is to mimick the proof of Gödel’s incompleteness theorem. Let $\pri(x)$ be the $\Sigma_3$ formula naturally expressing the predicate “the sentence $x$ is provable from $\pa$ using one application of the $\omega$-rule”. By formalizating in $\pa$ the easy arguments that (1) unnested applications of the $\omega$-rule can be collapsed to one, and (2) all true $\Sigma_3$ sentences are provable by one application of the $\omega$-rule, we obtain that $\pri$ satisfies the Hilbert–Bernays–Löb derivability conditions:


*

*$\pa\vdash_1 A\implies\pa\vdash_1\pri(\code A)$,

*$\pa\vdash\pri(\code{A\to B})\to(\pri(\code A)\to\pri(\code B))$,

*$\pa\vdash\pri(\code A)\to\pri(\code{\pri(\code A)})$,
where $\vdash_1$ denotes provability using one application of $\omega$-rule. Thus, by the standard proof of the second incompleteness theorem,
$$\pa\nvdash_1\neg\pri(\code\bot).$$
That is, if $A=\pri(\code\bot)$ is the sentence asserting the $\omega$-inconsistency of $\pa$, then $\neg A$ is not provable from $\pa$ using one application of $\omega$-rule, or in other words, $\pa+A$ is $\omega$-consistent. On the other hand, $\neg A$ is true, hence $\pa+\neg A$ is also $\omega$-consistent.
A: Another proof (due to G. Kreisel, [1]) :
Using diagonal lemma construct a sentence $K$ such that $$PA\vdash K \leftrightarrow \neg \omega -con(PA+K)  ~~~~~~(I)$$ As it was noticed in the Emil's proof, omega-inconsistency is a $\Sigma_3$ property, so $K$ is a $\Sigma_3$ sentence. $K$ is false in the standard model (because if it was true, then by $(I)$ , $PA+K$ should be omega-inconsistent which is impossible, because $\mathbb{N}\vDash PA+K$). So $\mathbb{N}\vDash \neg K$ and (again by $(I)$), $\mathbb{N}\vDash \omega -con(PA+K)$, therefore $PA+K$ is omega-consistent. But $PA+\neg K$ is also omega-consistent because  $\mathbb{N}\vDash \neg K$.
Reference
[1] "Necessary and sufficient conditions for undecidabillity of the Gödel sentence and its truth", Peter Clark, David DeVidi, and Michael Hallett (eds), Vintage Enthusiasms:  Essays in Honour of John Bell, University of Western Ontario
