How can we know arithmetical axioms are consistent? If we assume both distributivity and the opposite of the law of signs (ie, that $-1\times-1 = -1$) for the relative integers, then we can derive that two different numbers are actually equal.
$$-2(5+-3) = -2\times2 = -4$$
but,
$$-2(5+-3) = -2\times5 + -2\times-3 = -10 + -6 = -16$$
The axioms that are conventionally assumed for the integers are simply the ring axioms. My question is, if the set of "axioms" described above turns out to be inconsistent, how can we be so sure that the ring axioms aren't inconsistent as well?
I'm aware of this question. On the one hand, I'm asking if you really need to venture that deeply into proof theory just for this (seemingly simple) special case. But if you do, could you provide an example of how to apply that technique to prove this special case?

The ring axioms are consistent because we can form models of these axioms; this is the content of the so-called soundness property of first-order logic: is a set of axioms has a model, then it is consistent.  For an easy example (noted by Henning Makholm in his comment), we can construct the two-element ring without extra assumptions, and show that it satisfies the ring axioms; of course, this ring is usually denoted $\mathbb Z / 2 \mathbb Z$, but we do not have to first construct the integers to create this ring.  
(The opposite property, that every consistent set of axioms has a model, is the heart of Gödel's Completeness Theorem.)
On the other hand, the axiom system you have devised is also consistent: it will also hold in $\mathbb{Z} / 2 \mathbb{Z}$.  All you have noticed is that $\mathbb Z$ is not a model of this set of axioms.
