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Since matrix mutiplication is not commutative, the two ways in which you can factorize matrices makes a difference in which side the factor goes on.

In particular, if I want to distribute

$$((I - A) + A)(I - A)^-1,$$

would it become

$$(I - A)(I - A)^-1 + A(I - A)^-1 $$

OR would it be

$$(I - A)^-1(I - A) + (I - A)^-1A?$$

How do I know which side it goes on? I think the first one is correct.

Your first answer is correct. There are two distributive laws for matrices,
$$A(B+C)=AB+ACquadhboxandquad (A+B)C=AC+BC ,$$
but not $A(B+C)=BA+CA$ or $(A+B)C=AC+CB$ or.....

In general, this is what we call "right distributivity" - I usually hear the context for this in the sense of ring axioms. Let's sojourn into this a bit - though if you're not familiar with abstract algebra, this won't be particularly enlightening, and you might be better off skipping to the very end.

Let $(R,+,cdot,0,1)$ be a ring; then we call left-distributivity and define it by

$$a cdot (b+c) = acdot b + a cdot c$$

Similarly, right-distributivity is given by

$$(b+c)cdot a = bcdot a + ccdot a$$

Note: we are not guaranteed that $acdot b = bcdot a$ unless $R$ is a commutative ring.

In the context of matrices over rings, for which I reference Wikipedia, you can define $M_n(R)$ as the $ntimes n$ matrices over a ring $R$ (i.e. its elements come from the ring, and the addition and multiplication of elements are shared). Notably, we have that $M_n(R)$ is a commutative ring if and only if $R$ is a commutative ring and $n=1$ (so basically effectively no different from working in the ring in question).

So what does this mean? This means, in your case, you probably do not have $AB=BA$ (of course, I imagine you know this). And thus in the context of the distributivity thigns above, you would have

$$(B+C)A = BA + CA$$

Your example has $B = I-A$ and $C=A$. And thus, your first example is correct: if you are distributing something on the right side, and cannot ensure commutativity, you should multiply that element by everything in the brackets on the right side.