Fact 1 (Goldowsky-Tonelli): Let $F:(a, b) \to \mathbb{R}$ be continuous and have finite derivative everywhere. Suppose $F' \geq 0$ almost everywhere. Then $F$ is monotonically increasing.

For a proof of this, see Saks, Theory of the integral, Chapter 6, page 206.

Suppose $X = \{x \in [a, b]: -1 < f'(x) < 1\}$ has zero measure. Let $Y = \{x \in [a, b]: f'(x) \leq -1\}$ and $Z = \{x \in [a, b]: f'(x) \geq 1\}$

Claim 1: Every point of $X$ is a limit point of $Y$ and a limit point of $Z$.

Proof: Suppose for example $x \in X$ is not a limit point of $Y$ - the other case is similar. Let $I$ be an open interval around $x$ disjoint with $Y$. Then at almost every $y \in I$, $f'(y) \geq 1$. Using Fact 1, it follows that the function $y \mapsto f(y) - y$ is monotonically increasing on $I$ and hence for every $y \in I$, $f'(y) \geq 1$ which is impossible as $x \in Y \cap X$.

Claim 2: The set of points of continuity of $f' \upharpoonright \overline{X}$ is dense in $\overline{X}$ (the closure of $X$).

Proof: Well known (using Baire category theorem).

Now let $I$ be any open interval around $x \in X$. Then the supremum of $f' \upharpoonright I$ is at least $1$ and its infimum is at most $-1$ by Claim 1. By Darboux theorem, it follows that $f' \upharpoonright I$ and hence also $f' \upharpoonright (I \cap X)$ takes every value in $(-1, 1)$. So $f'$ is everywhere discontinuous on $\overline{X}$ which contradicts Claim 2.

Real and Complex Analysis) the fundamental theorem of calculus holds, so $f(x) - f(a) = \int_a^x f'(t)\,dt = x - a$, i.e., $f(x)$ is actually the linear function $x-a+f(a)$ and in particular $f'(b) = 1 \ne -1$. So I expect the answer to be no in general because I don't think it can help you to allow $f'(x)$ to exceed 1. $\endgroup$ – Timothy Chow Apr 4 '17 at 21:00