I have already posted an answer but in light of the discussion and the kinds of confusions that have emerged, I believe that this additional answer will be helpful.
Let us first note that the consistency of PA, or more precisely a certain formalized version of it that I will call "Con(PA)," is a theorem of Zermelo-Fraenkel set theory (ZF). Conceptually, the simplest ZF proof is obtained by formalizing the easy and almost trivial argument that N, the natural numbers, is a model of PA.
ZF is an extremely powerful system, and the full power of ZF is not needed for proving Con(PA). Famously, Gentzen showed that primitive recursive arithmetic (PRA), a very weak system, can prove Con(PA) if you add the ability to do induction up to the countable ordinal $\epsilon_0$. Other ways to prove Con(PA) are available. Let B-W denote the statement that "every bounded sequence of rational numbers contains a subsequence $(q_i)$ such that for all $n$, $|q_i - q_j| < 1/n$ for all $i, j > n$." Then B-W implies Con(PA), and this implication can certainly be proven in the system RCA$_0$, and (according to Harvey Friedman) actually much less than RCA$_0$ is needed.
Most mathematical statements are no longer considered "open problems" once a proof has been published or otherwise made widely available, and checked and confirmed by experts to be correct. Note that published proofs, and expert verification, usually make no explicit reference to any particular underlying formal system such as ZF or PRA. Mathematicians are trained to recognize correct proofs when they see them, even if no set of axioms is explicitly specified. If pressed to specify an axiomatic system, a common choice is ZF, or ZFC (ZF plus the axiom of choice). If a proof is available that is explicitly formalizable in ZF, that is normally regarded as more than sufficient for settling an assertion and removing its "open problem" status.
In the case of Con(PA), the aforementioned "normal conditions" for removing its "open problem" status have been met, and in fact exceeded. Nevertheless, some debate continues over its status, most likely because Con(PA) is widely perceived to be a somewhat unusual mathematical statement, having connections to philosophical questions in the foundations of mathematics. For example, some people, whom I will loosely call "formalists" or "ultrafinitists," maintain that many ordinary mathematical statements (e.g., "every differentiable function is continuous") have no concrete meaning, and the only concrete thing that can be said about them is whether they can or cannot be proved in this or that formal system; however, a statement such as "PA is consistent" is regarded as having a direct, concrete meaning. Roughly speaking this is because "PA is inconsistent," unlike infinitary mathematical statements, can be assigned a quasi-physical meaning as the existence of a certain finite sequence of symbols that we can physically apprehend. While the formalist agrees with all the above facts about the provability of Con(PA) in this or that formal system, such formal proofs don't necessarily carry any weight with the formalist as far as establishing the consistency of PA (in what I've called the "quasi-physical" sense) goes. Formalists will generally agree that explicitly exhibiting a contradiction in PA will definitively establish its inconsistency, but may differ regarding what, if anything, would definitively establish its consistency.
There are others who are not formalists but who reject the commonly accepted standard of ZF(C) and only accept proofs that are formalizable in much weaker systems. For example, someone with strong constructivist leanings might only accept proofs that are formalizable in RCA$_0$. For such a person, the proof of Con(PA) in ZF carries no weight. Roughly speaking, the usual ZF proof, that proceeds by showing that N is a model of the axioms of PA, assumes that any first-order formula defines a set of natural numbers, and this assumption is unprovable on the basis of RCA$_0$ alone. In fact, one can prove that Con(PA) is unprovable in RCA$_0$. Such a person might regard the consistency of PA as permanently unknowable (in a way similar to those who regard the continuum hypothesis as permanently unknowable since it has been proved independent of ZFC). Note, by the way, that this person would also regard a sizable portion of generally-accepted mathematics (including Brouwer's fixed-point theorem, the Bolzano-Weierstrass theorem, etc.) as being "unproved" or "unprovable."
To summarize, the consistency of PA is not an open problem in the usual sense of the term "open problem." Some people do nevertheless assert that it is an open problem, or that it has not been proven. When you encounter such an assertion, you should be aware that most likely, the person is using the term "open problem" in a somewhat nonstandard fashion, and/or holds to certain standards of proof that are more stringent than those that are commonly accepted in the mathematical community.
Finally, to answer the new question that Franklin has asked, about whether the consistency of any of the systems in which Con(PA) has been proved has been proved: The answer is, "not in any sense that you would likely find satisfying." For example, one can "prove" that PRA + induction up to $\epsilon_0$ is consistent, in the sense that the consistency proof can be formalized in ZF, which as I said above is the usual standard for settling mathematical questions. If, however, the reason that you're asking the question is that you doubt the consistency of PA, and are hoping that you can settle those doubts by proving the consistency of PA in some "weaker" system that can then be proved consistent using "weaker" assumptions that you don't have any doubts about, then you're basically out of luck. This, roughly speaking, was Hilbert's program for eliminating doubts about the consistency of infinitary set theory. The hope was that one could prove the consistency of (say) ZF on the basis of a weak system such as (say) PRA, about which we had no doubts. But Goedel showed that not only is this impossible, but even if we allow all of ZF into our arsenal, we still can't prove the consistency of ZF. For better or for worse, this tempting road out of skepticism about consistency is intrinsically blocked.