I will give two answers:
Do it without complex numbers, notice that
$$ \begin{eqnarray}
\mathcal{F}(\omega) = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x &=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x + \int_{-\infty}^0 \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x \\
&=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x + \int_{0}^{\infty} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{-j \omega x} \mathrm{d} x \\
&=& 2 \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \cos(\omega x) \mathrm{d} x
\end{eqnarray}
$$
Now, compute $\mathcal{F}^\prime(\omega)$, and integrate by parts:
$$\begin{eqnarray}
\mathcal{F}^\prime(\omega) &=& -\frac{2}{\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{x^2}{2}} x \sin(\omega x) \mathrm{d} x = \frac{2}{\sqrt{2\pi}} \int_0^\infty \sin(\omega x) \mathrm{d} \left( \mathrm{e}^{-\frac{x^2}{2}} \right) \\
&=& \frac{2}{\sqrt{2\pi}} \left. \mathrm{e}^{-\frac{x^2}{2}} \sin(\omega x) \right|_0^\infty - \frac{2}{\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{x^2}{2}} \omega \cos(\omega x) \mathrm{d} x \\
&=& - \omega \mathcal{F}(\omega)
\end{eqnarray}
$$
The solution to so obtained ODE, $\mathcal{F}^\prime(\omega) = - \omega \mathcal{F}(\omega)$ is $\mathcal{F}(\omega) = c \exp\left( - \frac{\omega^2}{2} \right)$, and the integration constant is seen to be one from normalization requirement $\mathcal{F}(0)=1$ of the Gaussian probability density.
Complex integration: As you have started, complete the square:
$$ \left( -\frac{x^2}{2} + j \omega x \right) =
\left( -\frac{x^2}{2} + j \omega x + \frac{\omega^2}{2} \right) - \frac{\omega^2}{2} =
-\frac{1}{2} \left( x - j \omega \right)^2 - \frac{\omega^2}{2}
$$
We then have:
$$
\mathcal{F}(\omega) = \mathrm{e}^{-\frac{\omega^2}{2}} \cdot \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x
$$
The integral $I = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x$ is indeed $1$. To see this, consider
$$
\begin{eqnarray}
I_L &=& \int_{-L}^L \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x = \int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \\
&=& \left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) + \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \\
&=& \left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) + \mathcal{I}_L
\end{eqnarray}
$$
Here we denoted $\mathcal{I}_L = \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z$. Notice that $\lim\limits_{L \to \infty} \mathcal{I}_L = 1$.
Consider a complex contour $\mathcal{C}$, $ -L \to L \to L - j \omega \to -L - j \omega \to -L$:
$$
\begin{eqnarray}
I_L - \mathcal{I}_L &=&\left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) \\ &=& -\int_\mathcal{C} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{L}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L-j \omega}^{-L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z
\end{eqnarray}
$$
The integral over $\mathcal{C}$ is zero, since the integrand is holomorphic. Therefore:
$$
I-1 = \lim_{L \to \infty} (I_L-\mathcal{I}_L) = \lim_{L \to \infty} \left( - \int_{L}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L-j \omega}^{-L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \right)
$$
And the limit above is easily seen to vanish. Indeed:
$$
\lim_{L\to\infty} \left| \mathrm{e}^{-\frac{-(-L - j \omega t)^2}{2}} \right| =
\lim_{L\to\infty} \left| \mathrm{e}^{-\frac{-(L^2 + \omega^2 t^2)}{2}} \right| =0.
$$