I'm reading in my notes that $\mathcal{L}\left( h\left( t \right) x\left( t \right) \right) = \left( H * X \right)\left( s \right)$.
[$*$ is the convolution]
I'm surprised not to see the $\frac{1}{2 \cdot \pi}$ factor in the Laplace transform (as we have it in Fourier transform), is that a mistake in my notes?
The $2\pi$ in the Fourier transform appears because of its relationship to the Fourier series (and, consequently, periodic functions). While you can get Laplace from Fourier by making a suitable variable change, we should note that the $2\pi$ has a slightly different meaning here.
If $f$ admits a Laplace transform $F$, then the way to formally recover $f$ is by the integral
$$ f(t) = \frac{1}{2\pi i} \int_{c - i\infty} ^{c + i\infty} e^{st} F(s) \ ds $$
where the integral ranges over a line in the complex plane containing the poles of the integrand to its right. The $2\pi$ here has nothing to do with its relationship in the Fourier transform, but is here because of the residue theorem.