Frequency shift property of Fourier transform
Your Eq. $(2)$ is the frequency shift property; Eq. $(1)$ is wrong. This is easy to show:

$$\begin{align}\mathcal{F}\left\{e^{j\omega_0t}x(t)\right\}=&amp;\int_{-\infty}^{\infty}x(t)e^{j\omega_0t}e^{-j\omega t}dt\\&amp;=\int_{-\infty}^{\infty}x(t)e^{-j(\omega-\omega_0)t}dt\\&amp;=X(j(\omega-\omega_0))
\end{align}$$

Your 'Case II' has nothing to do with the frequency shift property, because it is no frequency shift. For a frequency shift you need to multiply by a complex exponential $e^{j\omega_0t}$, and not by a real-valued exponential $e^{at}$.

You can come up with a rule for the Fourier transform of $x(t)e^{-at}$, $a\in\mathbb{R}$, but this is more tricky than the frequency shift property. For the frequency shift property, if you know that $X(j\omega)$ exists, then you know for sure that also the Fourier transform of $x(t)e^{j\omega_0t}$ exists. On the other hand, if $X(j\omega)$ exists, it is not certain that also the Fourier transform of $x(t)e^{-at}$ exists. This has everything to do with the Laplace transform and its region of convergence:

$$\begin{align}\mathcal{F}\left\{x(t)e^{-at}\right\}&amp;=\int_{-\infty}^{\infty}x(t)e^{-at}e^{-j\omega t}dt\\&amp;=\int_{-\infty}^{\infty}x(t)e^{-(a+j\omega)t}dt\stackrel{?}{=}X(a+j\omega)\end{align}$$

The last equality only holds if the integral converges, i.e. if $s=a$ is inside the region of convergence of the Laplace transform of $x(t)$, and if the Laplace transform of $x(t)$ has no singularities on the imaginary axis, i.e. if its Fourier transform has no Dirac delta impulses. The latter explains why the above 'rule' doesn't work for $x(t)=u(t)$ (as in your example).