I would close the contour in the upper half of the complex plane, the principal value picks up $i\pi$ times the residue$^\ast$ at $t=0$, which is $u/(1-u)$. There are no other poles.$^{\ast\ast}$

$^\ast$ $\frac{1-e^{i t u}}{e^{i t u}-i t-1}=\frac{u}{1-u}+{\cal O}(t^2).$

$^{\ast\ast}$ poles are at $t=i\tau$ with $e^{-\tau u}+\tau=1$ (excluding $\tau=0$, which is canceled by the numerator); these remain at $\tau<0$ for all $u\in(0,1)$, approaching $-2(1-u)$ for $u\rightarrow 1$.

In the comments there was an issue with the numerical evaluation. Principal value integrals of this type can be evaluated more accurately by replacing $1/t$ by $\frac{d\log |t|}{dt}$ and carrying out a partial integration. This gives
$$\int_{-\infty}^\infty dt\,\frac{1-e^{itu}}{e^{itu}-1-it}\,\frac{1}t=
-2i\Im\int_{0}^\infty dt\,\ln|t|\frac{d}{dt}\frac{1-e^{itu}}{e^{itu}-1-it}.$$
For the case $u=1/2$ considered in the comments, Mathematica gives 3.1406.

`NIntegrate[(1 - Exp[I*t*1/2])/(Exp[I*t*1/2] - 1 - I*1/2)/ t, {t, -1000, -0.001}, AccuracyGoal -> 3, PrecisionGoal -> 3] + NIntegrate[(1 - Exp[I*t*1/2])/(Exp[I*t*1/2] - 1 - I*1/2)/t, {t, 0.001, 1000}, AccuracyGoal -> 3, PrecisionGoal -> 3]`

produces $1.25389\, +2.51409 i$ and does not confirm your hypothesis. $\endgroup$ – user64494 Nov 20 at 19:07`NIntegrate[(1 - Exp[I*t*1/2])/(Exp[I*t*1/2] - 1 - I*1/2)/ t, {t, -10000, -0.0001}, AccuracyGoal -> 3, PrecisionGoal -> 3, WorkingPrecision -> 30] + NIntegrate[(1 - Exp[I*t*1/2])/(Exp[I*t*1/2] - 1 - I*1/2)/t, {t, 0.0001, 10000}, AccuracyGoal -> 3, PrecisionGoal -> 3, WorkingPrecision -> 30]`

produces $1.2564281632324901625528374684+2.51331913735615084972161764584 i$. $\endgroup$ – user64494 Nov 20 at 19:31