Can a Fourier Transform exist even if the j$\omega$ axis is not in the Region of Convergence in it's Laplace Transform
You're right that the Laplace transform is not more general than the Fourier transform. They are just different. There are several (theoretically) important functions for which the Laplace transform doesn't exist, but the Fourier transform does. A few examples are


$x(t)=e^{j\omega_0t}$
$x(t)=\sin(\omega_0t+\phi)$
$x(t)= \textrm{sinc}(\omega_0t)$
$x(t)=\frac{1}{\pi t}$
$x(t)=\textrm{sign}(t)$


The first two involve Dirac impulses in their Fourier transform, whereas the third and the fourth have discontinuous Fourier transforms. I hope it is obvious why the complex exponential and the sine are important. The $\textrm{sinc}$ function is needed to represent the impulse response of ideal frequency-selective filters, such as low pass and high pass filters, and $1/(\pi t)$ is the impulse response of an ideal Hilbert transformer. Note that those functions are non-causal (i.e., they are not zero for $t&lt;0$). The class of non-causal functions for which the (two-sided) Laplace transform exists is quite restricted because damping through multiplication with $e^{st}$ only works either for $t&gt;0$ (if $\textrm{Re}\{s\}&lt;0$) or for $t&lt;0$ (if $\textrm{Re}\{s\}&gt;0$), but not for both.

In your first example ($\sin(t)u(t)$) you got the Fourier transform wrong, even though it does contain Dirac impulses. The Fourier transform has Dirac impulses whenever the Laplace transform (if it exists) has poles on the imaginary axis.