Does an elementary method exist?

It depends on what you call elementary. Here is a solution without contour integration.

The function $f(x) \,=\, e^{-|x|}$ has Fourier transform $(\mathcal{F}f)(t) \; = \; \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} f(x)e^{-ixt}\,dx \; = \; \sqrt{\frac{2}{\pi}} \frac{1}{1+t^2}$ and so, since the integrand is even, $\int_{\mathbb{R}} \frac{\cos x}{x^2+1}\,dx \; = \; \sqrt{\frac{\pi}{2}}\int_{\mathbb{R}}(\mathcal{F}f)(x)e^{ix}\,dx \; = \; \pi f(1) \; = \; \frac{\pi}{e}$ Thus the integral we want is $\int_{\mathbb{R}}\frac{\sin x}{x^2 + 4x + 5}\,dx \; = \; \int_{\mathbb{R}}\frac{\sin(x-2)}{x^2+1}\,dx \; = \; -\sin 2\int_{\mathbb{R}}\frac{\cos x}{x^2+1}\,dx \; = \; -\pi e^{-1} \sin 2$ making the answer $1 \times -1 \times 2 = \boxed{-2}$ .

$\begin{aligned} \int_{-\infty}^\infty \dfrac{ \sin x}{x^2+4x+ 5} \, dx &=& \int_{-\infty}^\infty \dfrac{ \sin x}{(x+2)^2+1} \, dx \stackrel{y=x+2}{\LARGE =} \int_{-\infty}^\infty \dfrac{ \sin (y-2)}{y^2+1} \, dy \\ &=& \int_{-\infty}^{\infty} \dfrac{ \sin y \cos 2 - \cos y \sin 2}{y^2+1} \, dy \\ &=& \cos 2 \underbrace{ \int_{-\infty}^\infty \dfrac{\sin y}{y^2+2} \, dy }_{=0 \; (\because \text{ odd function})}- \sin 2 \int_{-\infty}^\infty \dfrac{\cos y}{y^2 + 1} \, dy \\ &=&- \sin 2 \int_{-\infty}^\infty \dfrac{\cos y}{y^2 + 1} \, dy \end{aligned}$

Claim : $\displaystyle \int_{-\infty}^\infty \dfrac{\cos y}{y^2 + 1} \, dy = \dfrac \pi e$ .

Proof : Consider the Laplace transform of the function $\displaystyle g(a) = \int_0^\infty \dfrac{ \cos(ax) }{1+x^2} \, dx$ ,

$\mathcal L (g(a)) = \int_0^\infty \int_0^\infty \dfrac{ \cos(ax)}{1+x^2} e^{-as} \, da \; dx = \int_0^\infty \dfrac s{(1+x^2)(s^2+x^2) } \, dx = \dfrac \pi{2(s+1)} \; .$

Then $\displaystyle g(a) = \mathcal L^{-1} \left( \dfrac \pi{2(s+1)} \right) =\dfrac \pi 2 e^{-|a|} \Rightarrow \int_{-\infty}^\infty \dfrac{ \cos(x) }{1+x^2} \, dx = 2g(1) = \dfrac \pi e$ .

Hence, $\displaystyle \int_{-\infty}^\infty \dfrac{ \sin x}{x^2+4x+ 5} \, dx = - \sin 2 \cdot \dfrac \pi e$ .