The solution is $$f(y)=\frac{2}{\pi}\frac{y}{1+y^2}$$ I will add the derivation soon.
Derivation:
Let
$$g(y)=
\begin{cases}
f(y)&&y\ge0 \\
0&&y<0
\end{cases}
$$
Then, with Euler's formula, we can rephrase the problem as
$$\int^\infty_{-\infty}g(y)\left(\frac{e^{ixy}-e^{-ixy}}{2i}\right)dy=e^{-x}$$
Equivalently,
$$\mathcal F\{g(y)\}(-x)-\mathcal F\{g(y)\}(x)=i\sqrt{\frac2\pi}e^{-x}$$
$$G(-x)=G(x)+i\sqrt{\frac2\pi}e^{-x}\qquad{x>0}$$
This functional equation does not tell much, as the solution is not unique. The best we can do is defining
$$G(x)=
\begin{cases}
\varphi (x) && x>0 \\
\varphi(-x)+i\sqrt{\frac2\pi}e^{x} && x<0
\end{cases}
$$
for some $\varphi (x):\mathbb R^+$ with sufficiently nice properties.
Then,
$$
\begin{align}
g(y)&=\frac{1}{\sqrt{2\pi}}\int^\infty_{-\infty}G(x)e^{ixy}dx \\
\sqrt{2\pi}g(y)&=\int^\infty_{0}G(x)e^{ixy}dx+\int^{\infty}_{0}G(-x)e^{-ixy}dx \\
&=\int^\infty_{0}\varphi(x)e^{ixy}dx+\int^{\infty}_{0}\left(\varphi(x)+i\sqrt{\frac2\pi}e^{-x}\right)e^{-ixy}dx \\
&=2\int^\infty_{0}\varphi(x)\cos(xy)dx+i\sqrt{\frac2\pi}\int^{\infty}_{0}e^{-x}e^{-ixy}dx \\
&=2\int^\infty_{0}\varphi(x)\cos(xy)dx+i\sqrt{\frac2\pi}\frac1{1+iy} \\
g(x)&=\sqrt{\frac2\pi}\int^\infty_{0}\varphi(x)\cos(xy)dx+\frac i\pi\frac1{1+iy} \\
\end{align}
$$
Let $a>0$. By assumption, $g(-a)=0$. Therefore,
$$\sqrt{\frac2\pi}\int^\infty_{0}\varphi(x)\cos(ay)dx+\frac i\pi\frac1{1-ia}=0$$
Then,
$$\begin{align}
g(a)&=\sqrt{\frac2\pi}\int^\infty_{0}\varphi(x)\cos(ay)dx+\frac i\pi\frac1{1+ia} \\
&=-\frac i\pi\frac1{1-ia}+\frac i\pi\frac1{1+ia} \\
&=\frac2\pi\frac{a}{1+a^2}
\end{align}
$$
Hence, $$f(y)=\frac{2}{\pi}\frac{y}{1+y^2}$$
Verification:
(The calculations below are a bit sloppy as I assumed that the integral and differentiation can be interchanged.)
$$\begin{align}
L.H.S.
&=\frac2\pi \int^\infty_0\frac{y}{1+y^2}\sin(xy)dy \\
&=-\frac2\pi\frac{\partial}{\partial x}\int^\infty_0\frac{\cos(xy)}{1+y^2}dy \\
&=-\frac1\pi\frac{\partial}{\partial x}\int^\infty_{-\infty}\frac{\cos(xy)}{1+y^2}dy \\
&=-\frac1\pi\frac{\partial}{\partial x}\pi e^{-x} \\
&=e^{-x}\\
&=R.H.S.
\end{align}
$$
Here we utilized the well-known integral identity
$$\int^\infty_{-\infty}\frac{\cos(ax)}{1+x^2}dx=\pi e^{-|a|}$$
For its proof, see the accepted answer here.
Let $$F(w) = \int_0^\infty t^{-1/2} \sin (w t)\, dt$$
$$F(w) = w^{-1/2} \int_0^{\infty} w^{-1/2}t^{-1/2} \sin wt \, w dt$$
Let $q=wt$.
$$F(w) = w^{-1/2} \int_0^\infty q^{-1/2} \sin q \, dq = \text{const.}\cdot w^{-1/2}.$$
It is known that $\int_0^\infty q^{-1/2} \sin q \, dq = \sqrt{\frac{\pi}{2}}$, but we don't need to evaluate this integral to show that $t^{-1/2}$ is its own Fourier transform (up-to-a-constant).
UPDATE:
One way to compute $\int_0^\infty q^{-1/2} \sin q\, dq = 2\int_0^\infty \sin p^2 \, dp, $ is to consider the contour integral
$$ \oint_C e^{i z^2} \, dz$$ with C the segment $0$ to $R$ along the $x$-axis, a quarter circle of radius $R$ from $R$ to $z^\star=Re^{i\pi/4}$ and a segment from $z^\star$ to the origin (a slice (1/8) of a pie of radius $R$).
Then $$0=\oint_C e^{i z^2} \, dz = \int_0^\infty e^{ip^2} \, dp - e^{i\pi/4} \int_0^\infty e^{-y^2} \, dy + \int_{\Gamma_R} \exp (Re^{i2\theta})R i e^{i\theta}\,d\theta,$$ and
$$ \begin{aligned} \left| \int_{\Gamma_R} \exp (Re^{i2\theta})R i e^{i\theta}\,d\theta \right| &\le \int_{\Gamma_R} \left| \exp (R^2e^{i2\theta})R i e^{i\theta}\right| \,d\theta \\ &= \int_0^{\pi/4} Re^{-R^2\sin 2\theta} \,d\theta \\&<\int_0^{\pi/4} Re^{- \frac{4}{\pi} R^2\theta} \, d\theta \\&=\left(1-e^{-R^2} \right)\frac{\pi}{4R} \to 0.\end{aligned} $$
So
$$ \int_0^\infty \cos p^2 \, dp = \frac{1}{2} \sqrt{\frac{\pi}{2}},$$
$$ \int_0^\infty \sin p^2 \, dp = \frac{1}{2} \sqrt{\frac{\pi}{2}},$$
and the Fourier sine transform of $t^{-1/2}$ is $w^{-1/2}$.
$$t^{-1/2} \iff w^{-1/2}.$$
Best Answer
Hint: you can use complex numbers and write
$$e^{ikx}=\cos(kx)+i\sin(kx)$$
Calculating the integral of $e^{-x}e^{ikx}$ should be trivial. The real part of the answer is the cosine transform, the imaginary part is the sine transform.