As long as you axiomatize set theory in first-order logic, the answer to your question is no. The axioms would be consistent with each finite subset of the following set of sentences involving a new constant symbol $c$: "$c$ is a natural number" and "$c\neq n$" for each (standard name of a) natural number $n$. By compactness, there would be a model of the axioms plus all of these sentences, and in that model $c$ would denote a nonstandard natural number.
On the other hand, if you're willing to go beyond first-order logic, then the answer to your question is yes. For example, in second-order logic, you can express the induction axiom as a single sentence and be confident that "set" really means arbitrary set (not "internal set" or anything like that). In other words, once you're sure that "set" has its intended meaning, the induction principle guarantees that "natural number" also has its intended meaning. (To me, this doesn't look very helpful, since the intended meaning of "set" seems more complicated than the intended meaning of "natural number".)
For another example, if you're willing to use infinitary logic, then you can formulate the axiom "every natural number is equal to 0 or to 1 or to 2 or to 3, or ..."
The answer is no. It is enough to find a model of MA which is an integral domain of characteristic $0$ (whence O1 is true and E1 false) such that $2$ is not invertible (whence E2 is true and O2 false).
One example of such a model is the ring of $2$-adic integers $\mathbb Z_2$. This is clearly a domain, and $2$ is not a unit, hence it suffices to show
Theorem: For any prime $p$, the ring $\mathbb Z_p$ is a model of MA.
Proof: The only problem is to verify that induction holds. Assume $\mathbb Z_p\models\phi(0)\land\forall x\,(\phi(x)\to\phi(x+1))$, where $\phi$ is an arithmetic formula with parameters from $\mathbb Z_p$, and put $\phi(\mathbb Z_p):=\{a\in\mathbb Z_p:\mathbb Z_p\models\phi(a)\}$.
Since $\phi(\mathbb Z_p)$ is definable in $\mathbb Z_p$, it is also definable in the field $\mathbb Q_p$ endowed with a unary predicate for $\mathbb Z_p$. Macintyre [1] proved that such structures admit a form of quantifier elimination, and as a corollary (Thm. 2 on p. 609), every infinite definable set has a nonempty interior. Thus, there is $a_0\in\phi(\mathbb Z_p)$ and $k\ge0$ such that $a_0+p^k\mathbb Z_p\subseteq\phi(\mathbb Z_p)$. Let $a\in\mathbb Z_p$ be arbitrary, and let $b< p^k$ be a natural number such that $b\equiv a-a_0\pmod{p^k}$. Since $\phi(\mathbb Z_p)$ is closed under successor, we have $a\in b+a_0+p^k\mathbb Z_p\subseteq\phi(\mathbb Z_p)$. Thus, $\phi(\mathbb Z_p)=\mathbb Z_p$, i.e., $\mathbb Z_p\models\forall x\,\phi(x)$. QED
I suspect the following may work as additional countermodels (they are domains where $2$ is not a unit, the issue is whether they satisfy induction):
The ring of algebraic integers $\tilde{\mathbb Z}$. A form of quantifier elimination for $\tilde{\mathbb Z}$ was proved by van den Dries [2] and Prestel and Schmid [3], but the basic formulas are somewhat messy, so it is not immediately clear to me whether this implies induction.
The localization of $\tilde{\mathbb Z}$ at a maximal ideal containing $2$. Elimination of quantifiers for this (and similar) rings is reported as Fact 3 in [2], where it is attributed to [4]. It seems it could imply induction by a similar argument as for $\mathbb Z_p$.
[1] Angus Macintyre, On definable subsets of $p$-adic fields, Journal of Symbolic Logic 41 (1976), no. 3, pp. 605–610.
[2] Lou van den Dries, Elimination theory for the ring of algebraic integers, Journal für die reine und angewandte Mathematik 388 (1988), pp. 189–205.
[3] A. Prestel and J. Schmid, Existentially closed domains with radical relations, Journal für die reine und angewandte Mathematik 407 (1990), pp. 178–201.
[4] Angus Macintyre, Kenneth McKenna, Lou van den Dries, Elimination of quantifiers in algebraic structures, Advances in Mathematics 47 (1983), no. 1, pp. 74–87.
Best Answer
This is an answer to the edited questions Russell has added. Joel David Hamkins' reply in comments to the question is completely correct, but I'll take advantage of the greater space here.
Let (R,0,1,+,*) be a ring. Define
Sx = x + 1 and
B = {x | $\forall P(P0 \land \forall y\forall z(Py \land Sy,z \to Pz) \to Px)$}
i.e. B is the set of all x which are part of an S-chain beginning with 0.
Then S is functional and induction holds over B, i.e.
$\forall P(P0 \land \forall y\forall z(Py \land Sy,z \to Pz) \to \forall x (Bx \to Px))$.
One can define ++ and ** (both definitions being on B) from S using the normal recursive definitions, both of which can be proved commutative. By induction over B, one can show that + and ++ define the same function on B; also for * and **. Hence, if B = R, then * is commutative. So if R is a non-commutative ring, then B is properly contained in R.
"Can we prove induction fails in every non-commutative ring?" No. There are definitions of the successor function (see my other answer) so that induction will hold. OTOH, in a non-commutative ring R where the successor is defined as Sx = x + 1, induction will fail, because the set B (as defined above) cannot equal all of R. To see that induction fails, consider the predicate phi(n) to be Bn. Then clearly phi(0) and the inductive step holds, but obviously not phi(n) for all n in R.
"Is it impossible to define a successor chain that visits every ring element using addition in a non-commutative ring?" I'm not sure what you mean by this question. If successoring is defined by Sx = x + 1, then the successor chain isn't defined, it's implied (by the definition). You won't be able to prove that the successor chain is the entire ring, because again that would imply that multiplication is commutative, contrary to assumption.