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.
HINT
Firstly,
$$\begin{aligned}
&G(\omega,a,B)=a\int\limits_{-\infty}^\infty e^{-a^2B\left(\sqrt{\left(\frac ta\right)^2+1}-1\right)^2}e^{-ja\omega\frac ta}\,\mathrm d\dfrac ta=a\int\limits_{-\infty}^\infty e^{-a^2B(\sqrt{t^2+1}-1)^2}e^{-ja\omega t}\,\mathrm dt,\\
&G(\omega,a,B)=aH(a\omega,a^2B),\\
\end{aligned}$$
$$H(\Omega,c)=\int\limits_{-\infty}^\infty e^{-c(\sqrt{t^2+1}-1)^2}e^{-j\Omega t}\,\mathrm dt.\tag1$$
Let
$$\tau(t) = \sqrt{t^2+1}-1,\quad \tau\in[0,\infty),$$
so
$$t(\tau)=\pm\sqrt{(\tau+1)^2-1}=\pm\sqrt{\tau^2+2\tau},\quad t\in(-\infty,\infty),$$
$$dt=\pm\dfrac{\tau+1}{\sqrt{\tau^2+2\tau}},$$
$$\begin{aligned}
&H(\Omega,c)=\int\limits_{-\infty}^{-1}e^{-c\tau^2(t)}e^{-j\Omega t}\,\mathrm dt
+\int\limits_{-1}^\infty e^{-c\tau^2(t)}e^{-j\Omega t}\,\mathrm dt\\
&=-\int\limits_{\infty}^0e^{-c\tau^2}e^{j\Omega\sqrt{\tau^2+2\tau}}\dfrac{\tau+1}{\sqrt{\tau^2+2\tau}}\,\mathrm d\tau
+\int\limits_{0}^\infty e^{-c\tau^2}e^{-j\Omega \sqrt{\tau^2+2\tau}}\dfrac{\tau+1}{\sqrt{\tau^2+2\tau}}\,\mathrm d\tau,
\end{aligned}$$
$$H(\Omega,c)=2\int\limits_0^{\infty} e^{-c\tau^2} \cos(\Omega\sqrt{\tau^2+2\tau})\dfrac{\tau+1}{\sqrt{\tau^2+2\tau}}\,\mathrm d\tau.\tag2$$
By parts:
$$\begin{aligned}
&H(\omega,c) = \dfrac2\Omega\int\limits_0^{\infty} e^{-c\tau^2}\,d\sin(\Omega\sqrt{\tau^2+2\tau})\\
&=-\dfrac2\Omega e^{-c\tau^2} \sin(\Omega\sqrt{\tau^2+2\tau})\bigg|_0^\infty
+\dfrac{4c}\Omega\int\limits_0^{\infty} \tau e^{-c\tau^2}
\sin(\Omega\sqrt{\tau^2+2\tau})\,d\tau,\end{aligned}$$
$$H(\Omega,c)=\dfrac{4c}\Omega\int\limits_0^{\infty} \tau e^{-c\tau^2}\sin(\Omega\sqrt{\tau^2+2\tau})\,d\tau.\tag3$$
I don't see how to obtain the closed form of the integral $(3).$
At the same time, for $c<<\Omega$ the exponent in the formula $(1)$ allows polynomial approximation.
If $c\gtrsim\Omega,$ then the integral $(3)$ can be presented as $1D$ series, using the series expansion for sine expression in the form of
$$\sin(\Omega\sqrt{x^2+2x}) = \sqrt{\dfrac x2}\Omega\left(2 + \dfrac{3-4\Omega^2}6 x + \dfrac{16 \Omega^4 - 120 \Omega^2 - 15}{240} x^2 + \dfrac{-64\Omega^6 + 1680\Omega^4 - 1260\Omega^2 + 315}{20160} x^3 + \dfrac{256\Omega^8 - 16128\Omega^6 + 90720\Omega^4 + 15120\Omega^2 - 14175}{2903040} x^4 + \dfrac{-1024\Omega^{10} + 126720\Omega^8 - 2217600\Omega^6 + 1663200\Omega^4 - 623700\Omega^2 + 1091475}{638668800}x^5 + \dots\right).
$$
(the asymptotic form of
$$\sin(\Omega\sqrt{x^2+2x}) = \Omega{\cos((x+1)\Omega)\left(-\dfrac1{2x} + \dfrac 1{2x^2} + \dfrac{\Omega^2-30}{48x^3} - \dfrac{\Omega^2 - 14}{16x^4} - \dfrac{\Omega^4 - 540 \Omega^2 + 5040}{3840x^5} + \dfrac{\Omega^4 - 220 \Omega^2 + 1584}{768x^6} + \dots\right)} + {\sin((x+1)\Omega)\left(1 - \dfrac{\Omega^2}{8x^2} + \dfrac{\Omega^2}{4x^3} + \dfrac{\Omega^2(\Omega^2-168)}{384 x^4} - \dfrac{\Omega^2(\Omega^2-72)}{96 x^5} - \dfrac{\Omega^2(\Omega^4 - 1320 \Omega^2 + 59400)}{46080 x^6} + \dfrac{\Omega^2(\Omega^4 - 520 \Omega^2 + 17160)}{7680 x^7} + \dots\right)}
$$ is not so usable)
and the integral
$$\int\limits_0^\infty x^d e^{-cx^2}\,\mathrm dx = \dfrac{\Gamma\left(\frac{d+1}2\right)}{2c^{\frac{d+1}2}}.$$
Length of the series can be selected, using the numeric experiments.
Best Answer
$$\hat{f}(\omega)=\int_{-\infty}^\infty f(x)e^{-j\omega x}dx=\int_{-1}^1e^{-x}e^{-j\omega x}dx=\int_{-1}^1e^{-(1+j\omega)x}dx$$
Your above definition of Fourier Transform is valid if you are assuming non unitary angular frequecy $\omega$ but there are two other types of frequency, check their definitions also. But i will proceed with your definition and using $j=\sqrt{-1}$. Solving the exponential integral easily we get $$=-\left[\frac{e^{-x(1+j\omega)}}{1+j\omega}\right]_{-1}^{1}$$ $$=-\frac{1}{1+j\omega}\left[e^{-(1+j\omega)}-e^{(1+j\omega)}\right]$$ Using the property $\sinh x=\frac{e^x-e^{-x}}{2}$ we get $$=-\frac{1}{1+j\omega}\left[-2\sinh(1+j\omega)\right]$$ $$=\boxed{\color{blue}{\frac{2\sinh(1+j\omega)}{1+j\omega}}}$$
Edit : This is the best closed form in which we can express the answer, but if you wish to separate real and imaginary parts then you need to express in those long expressions as you did. Unfortunately there can't be simple closed forms for real and imaginary parts expressions here.