The primary thing we gain by looking at the derivative is that since the derivative is constant, our polynomial cannot have repeated roots, so none of the factors are repeated. It does not give us information on how many factors there are.

As you point out, the polynomial factors if \(\displaystyle a=0\). But you can do better than that: by Fermat, we know \(\displaystyle \alpha^p = \alpha\) for all \(\displaystyle \alpha \in F_p\), hence every \(\displaystyle \alpha\) is a root, hence \(\displaystyle x^p - x = (x-0)(x-1) \cdots (x-(p-1)) = x(x-1) \cdots (x-(p-1))\).

Now let us look at the general polynomial \(\displaystyle x^p - x + a\). We know this polynomial has a root in some extension of \(\displaystyle F_p\) (it might just be \(\displaystyle F_p\) itself, as in the case \(\displaystyle a = 0\)). Let \(\displaystyle \alpha\) be such a root in this (minimal) extension. Then what can we say about \(\displaystyle \alpha + 1\)?

I claim that if a finite field of characteristic \(\displaystyle p\) has one root of \(\displaystyle x^p - x +a\), then it has all of them, hence \(\displaystyle x^p - x +a\) splits completely over any extension of \(\displaystyle F_p\) in which it has a root. Think about why this is true. Then think about how you can use Galois theory to show that \(\displaystyle x^p - x + a\) is either irreducible over a field of characterstic \(\displaystyle p\) or factors completely. Then all that remains to show is why \(\displaystyle x^p - x + a\) does not have a root if \(\displaystyle a \neq 0\).