We have the following theorem:
$\textbf{Theorem}$. Let $|a_n| \downarrow 0$. Then, if $\sum_{n=1}^{\infty}a_n\sin nx$ converges absolutely at some point $x_0$ which is not a multiple of $\pi$, then $\sum_{n=1}^{\infty}|a_n|<\infty$.
This theorem can be found in N. K. Bary's book: A Treatise on Trigonometric Series, Vol. II, page 334.
Next, your series can be rewritten as
$$
\sum_{n\in \mathbb{Z} \backslash \{0 \}} 2\frac{|\sin nx|}{|n|}=4\sum_{n=1}^{\infty} \frac{|\sin nx|}{n}.$$
Since $1/n \downarrow 0$, if the series converges absolutely at some $x_0$ which is not multiple of $\pi$, then we would obtain the convergence of the harmonic series, which is obviously false.
I recommend you having a look at the proof of the cited theorem so you can see the difference between it and your approach.
EDIT: Proof of theorem:
If $\sum |a_n \sin nx_0|<\infty$, then $\sum |a_n \sin^2 nx_0|<\infty$ and $\sum |a_{n-1} \sin^2 (n-1)x_0|<\infty$. But
\begin{align}
|a_n|\sin^2 nx_0 +|a_{n-1}|\sin^2(n-1)x_0 & \geq |a_n|(\sin^2 nx_0 +\sin^2 (n-1)x_0)\\
& = |a_n|((1-\cos 2nx_0)/2+(1-\cos 2(n-1)x_0)/2) \\
&= |a_n| (1-\cos x_0 \cos(2n-1)x_0) \geq |a_n|(1-|\cos x_0|).
\end{align}
Hence,
$$
\infty> \sum |a_n|(1-|\cos x_0|) = (1-|\cos x_0|)\sum |a_n|,
$$
which gives the desired result, since $x_0$ is not multiple of $\pi$.
Suppose $a \ne -\pi$ or $b \ne \pi$ and $a\ne b$. Then the function you are talking about must be discontinuous.
Suppose the series did converge absolutely for some $x$. That would mean
$$
\sum_n \left|\frac{e^{-ina}-e^{-inb}}{2\pi in}\right| < \infty.
$$
But that would force the uniform convergence of the Fourier series everywhere by the Weierstrass M-test. But uniform convergence would imply that the periodic extension of the limit function $\chi_{[a,b]}$ must be continuous everywhere, which only happens in the case that $a=-\pi$ and $b=\pi$, or $a=b$.
I'm not familiar with you text, but you should have some pointwise convergence theorem that shows the Fourier series converges to the mean of the left and right hand limits for your function.
Best Answer
You have \begin{align} |e^{-ina}-e^{-inb}|^2 &=(\cos (-na) - \cos (-nb))^2+(\sin (-nb)-\sin (-na))^2\\ \ \\ &=2-2\cos(na)\cos(nb)-2\sin(na)\sin(nb)\\ \ \\ &=2-2\cos n(a-b)\\ \ \\ &=2-2\cos 2\tfrac{n(a-b)}2\\ \ \\ &=2-2\left(\cos^2\tfrac{n(a-b)}2-\sin^2\tfrac{n(a-b)}2\right)\\ \ \\ &=2-2\left(1-2\sin^2\tfrac{n(a-b)}2 \right)\\ \ \\ &=4\sin^2\tfrac{n(a-b)}2. \end{align} So $$ |e^{-ina}-e^{-inb}|=2\left|\sin\tfrac{n(a-b)}2\right| $$