Not elementary (in the sense of not using complex analysis), but this is how I'd do it:
Let $f(z) = \frac{1}{z^s(z+1)},$ where $z^s$ denotes the "natural branch", i.e. choose $\phi \arg z \in (0,2\pi)$ and put $(re^{i\phi})^s = r^s e^{is\phi}$. Take a "keyhole contour" C:

and integrate $f$ along $C$ using the residue theorem:
$\int_C f(z)\,dz = 2\pi i \operatorname{Res}_{z=-1} f(z).$
Estimating $\int_\gamma f(z)\,dz$ and $\int_\Gamma f(z)\,dz$, we have
$\left| \int_\gamma f(z)\,dz \right| \le \frac{M}{r^{\mathrm{Re}(s)}} \cdot 2\pi r \to 0 \qquad\text{as }r\to0$ and $\left| \int_\Gamma f(z)\,dz \right| \le \frac{M}{R^{1+\mathrm{Re}(s)}} \cdot 2\pi R \to 0 \qquad\text{as }R\to\infty.$
(For the first estimate, we want $\mathrm{Re}(s) < 1$ and for the second $\mathrm{Re}(s) > 0$.
For the two remaining line segments, on the "top" segment we get the integral we're looking for as $r \to 0$ and $R\to\infty$. On the "bottom" segment, we get (remember the choice of branch) $-\int_0^\infty \frac{1}{x^s e^{2\pi i s} (1+x)}\,dx.$
Putting it all together: $(1-e^{-2\pi i s}) \int_0^\infty \frac{1}{x^s(1+x)}\,dx = 2\pi i \operatorname{Res}_{z=-1} f(z) = 2\pi i (-1)^{-s} = 2\pi i e^{-\pi i s}$ so $ \int_0^\infty \frac{1}{x^s(1+x)}\,dx = 2\pi i \frac{e^{-\pi i s}}{1-e^{-2\pi i s}} = \frac{\pi}{\sin s\pi}.$