Is arithmetic on the naturals $\omega$-consistent? Is there any first order theory of arithmetic of the natural numbers that is known to be $\omega$-consistent if it is consistent?
If yes, then how?
 A: The empty theory over the language of arithmetic is certainly $\omega$-consistent.
This theory has the property that whenever it proves some formula $\phi$, you can replace every atomic formula $t_1=t_2$ by "true" and ignore all of the quantifiers, and the resulting Boolean expression will then evaluate to true. (Intuitively this is because the one-element universe is a model, but you can verify it purely syntactically one inference rule at a time).
Therefore it is impossible for the theory to prove both $\neg\phi(0)$ and $(\exists x)\,\phi(x)$, since the two Boolean expressions they translate to are each other's negations.

(This same argument also works for full Peano Arithmetic minus the axiom stating that $0$ is not a successor).
A: Let me add a fact that doesn't answer your question literally as stated, but may be interesting to you and other readers of this question: if you accept the axioms of ZFC, then pretty much any theory of arithmetic is $\omega$-consistent. Namely,

Theorem (ZFC).  For any theory of arithmetic $\mathcal{T}$ such that $\mathbb{N} \vDash \mathcal{T}$, $\mathcal{T}$ is $\omega$-consistent.$^1$

Here "theory" means "set of axioms", and $\vDash$ is the satisfaction relation from model theory: $\mathcal{M} \vDash \mathcal{T}$ means that the structure $\mathcal{M}$ (in this case, natural numbers with addition, multiplication, etc.) satisfies all of the axioms $\mathcal{T}$.
Since pretty much any theory of $\mathbb{N}$ is supposed to start from true axioms, that is axioms we believe are actually true about $\mathbb{N}$, most such theories will be $\omega$-consistent. In particular:

Corollary (ZFC). Peano Arithmetic (PA) and Robinson Arithmetic (RA) are $\omega$-consistent.

This may be a bit surprising! So what is the basis for these results? The theorem requires a strong meta-theory, in this case ZFC, in which we can develop all of model theory (in particular, in which we can define what a formula is, what a theory is, and what satisfaction $\vDash$ means). In such a meta-theory, there is also an object called $\mathbb{N}$, the natural numbers (in ZFC the existence of $\mathbb{N}$ is given as an axiom). Because such an object exists, we can then show that any correct theory (one that consists of axioms which are true about $\mathbb{N}$) is $\omega$-consistent.
If we do not get to assume strong meta-theory axioms like ZFC, but instead are wondering about the $\omega$-consistency of a theory like PA in a much weaker theory, then we may no longer be able to show that PA is $\omega$-consistent.

$^1$ Proof of the theorem: We proceed by contradiction: let $P(x)$ be a formula with one free variable, such that $\mathcal{T}$ proves (implies) $P(0)$, $P(1)$, and so on, but $\mathcal{T}$ also proves (implies) $\exists x. \lnot P(x)$.
Now since $\mathbb{N} \vDash \mathcal{T}$, $\mathbb{N} \vDash \varphi$ for any logical consequence (implication) $\varphi$ of the axioms $\mathcal{T}$. Therefore,
$\mathbb{N} \vDash P(i)$ for all natural numbers $i \in \mathbb{N}$, and moreover, $\mathbb{N} \vDash \exists x. \lnot P(x)$. But these statements are in direct contradiction: by definition, the latter statement means that there exists some natural number $j$ such that $\mathbb{N} \vDash \lnot P(j)$, which is the exact opposite of the former statement. So we are done.
