Fourier transform of dirac comb with function: The scaling factor Multiplication in the time domain corresponds to convolution in the frequency domain: 

$$
f(t) \cdot x(t) \iff F(j \omega) * X( j \omega)  \tag*{No scaling factor} 
$$
I know the fourier transform of the dirac comb is: 
$$
\mathcal{F} \big \{ \text{III}_{T_{s}} (t) \big \} =  \omega_{s} \cdot \text{III}_{T_{s} } (j \omega)  
$$
But according to oppenheim and others: 

$$
\mathcal{F} \big \{ f(t) \cdot \text{III}_{T_{s}} (t) \} =  \underbrace{\dfrac{1}{T_{s}}}_{\text{scaling factor}}  \cdot \displaystyle \sum_{k \to - \infty}^{ k \to \infty} X(j( \omega - k \omega_{s} ))  \text{ (Scaling factor)} 
$$
So my question is, for the above, why does the scaling factor not remain $$ \omega_{s} $$ why does it become $$ \dfrac{1}{T_{s}} $$ 
Which means that 
$$
x(t) \cdot y(t) \iff \dfrac{1}{2 \pi}  X(j \omega) * Y(j \omega) \tag{ ??? } 
$$
Why does the scaling factor not remain the sampling angular velocity?
Your first formula is wrong. If you use angular frequency ($\omega$), then multiplication in the time domain corresponds to $1/(2\pi)$ times convolution in the frequency domain, just as in the last formula in your question:

$$\mathcal{F}\{x(t)\cdot y(t)\}=\frac{1}{2\pi}X(j\omega)\star Y(j\omega)\tag{1}$$

If you use frequency $f=\omega/(2\pi)$, then you get

$$\mathcal{F}\{u(t)\cdot v(t)\}=U(f)\star V(f)\tag{2}$$

where I've used the definitions

$$X(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{3}$$

$$U(f)=\int_{-\infty}^{\infty}u(t)e^{-j2\pi f t}dt\tag{4}$$