If $f'$ exists everywhere, then we know that it cannot have any simple discontinuities. But in this case we only know that $f'$ exists a.e. (since $f$ is absolutely continuous).
More specifically, I've also been assuming that $f'$ is increasing (on the set where it is defined). In this case, $f'(x+)$ exists everywhere, and I've been asking if there can be a point $x$ such that $f'(x)$ exists but $f'(x)<f'(x+)$. (We know this cannot happen if $f$ is differentiable on $[x, x+\epsilon)$.)
On
No, $f'(x)$ can not have simple discontinuities.
Theorem
Let $f(x)$ be differentiable on some interval $(a,b)$, and continuous on $[a,b]$.
If $\lim_{x \to a^+}f'(x)$ exists, then it is equal to $f'_+(a)$ (the right derivative), same for the left derivative. (I will prove this bellow)
Consequence:
Because a simple discontinuity at $c\in(a,b)$ implies existence of sideways limits $\lim_{x \to c^+}f'(x)$ and $\lim_{x \to c^-}f'(x)$, and because $f$ is differentiable at $c \ $ $\left(f'_+(c) = f'_-(c) = f'(c)\right)$, it follows from the theorem above that
$$\lim_{x \to c^+}f'(x) = f'_+(c) = f'(c) = f'_-(c) = \lim_{x \to c^-}f'(x)$$
And thus $f'$ cannot have a simple discontinuity.
Proof of theorem stated above: $$ f'_+(a) = \lim_{h \to 0^+}\frac{f(a + h) - f(a)}{h}, $$
Now if you apply the MVT on the interval $[a,a+h]$,
\begin{equation} \frac{f(a + h) - f(a)}{h} = f'(a +h\theta) \end{equation}
where $\theta \in (0,1)$
Because $\lim_{x \to a^+}f'(x) = L$ exists,
$$ \lim_{h \to 0^+}f'(a +h\theta) = L = \lim_{h \to 0^+}\frac{f(a + h) - f(a)}{h} = f'_+(a) $$
I believe that $f(x) = |x|$ is a counterexample. Derivative exists except at $x = 0$, but is discontinuous there.