Does convolution violate duality of a Fourier Transform
Try using this definition of the continuous Fourier Transform and its inverse:

$$ \mathscr{F} \Big\{ x(t) \Big\} \triangleq X(f) \triangleq \int\limits_{-\infty}^{\infty} x(t) \, e^{-j2 \pi f t} \ \mathrm{d}t $$

$$ \mathscr{F}^{-1} \Big\{ X(f) \Big\} \triangleq x(t) = \int\limits_{-\infty}^{\infty} X(f) \, e^{j2 \pi f t} \ \mathrm{d}f $$

see how well that works for your duality.

Okay, I'll be more explicit.  Like your "$\omega$" definition, we have linearity:

$$ \mathscr{F} \Big\{ x(t) + y(t) \Big\} = \mathscr{F} \Big\{ x(t) \Big\} + \mathscr{F} \Big\{ y(t) \Big\} = X(f) + Y(f) $$

and derivative:

$$ \mathscr{F} \Big\{ \tfrac{\mathrm{d}}{\mathrm{d}t} x(t) \Big\} = j2\pi f \cdot X(f) $$

and delay:

$$ \mathscr{F} \Big\{ x(t-\tau) \Big\} = e^{-j2\pi f \tau} \cdot X(f) $$

But we get duality as:

$$\mathscr{F} \Big\{ X(t) \Big\} = x(-f)$$
$$\mathscr{F}^{-1} \Big\{ x(f) \Big\} = X(-t)$$

We get for convolution:

$$ \mathscr{F} \Big\{ h(t) \circledast x(t) \Big\} = H(f) \cdot X(f) $$

and it's the same in reverse:

$$ \mathscr{F}^{-1} \Big\{ W(f) \circledast X(f) \Big\} = w(t) \cdot x(t) $$

The other cool thing about this unitary definition of the Fourier Transform is how Parseval's theorem come out:

$$ \int\limits_{-\infty}^{\infty} \big| x(t) \big|^2 \ \mathrm{d}t = \int\limits_{-\infty}^{\infty} \big| X(f) \big|^2 \ \mathrm{d}f $$

And the more general Plancherel's theorem:

$$ \int\limits_{-\infty}^{\infty} x(t) \big(y(t)\big)^* \ \mathrm{d}t = \int\limits_{-\infty}^{\infty}  X(f) \big(Y(f)\big)^* \ \mathrm{d}f $$

where $(\cdot)^*$ denotes the complex conjugate.

No funky scaling factors.

This is, in my opinion, clearly the most natural definition of the Fourier Transform if you are never relating it directly to the Laplace Transform.  If you are bopping between the F.T. and the L.T. maybe your definition is better, but then I would say "$X(j \omega)$" so that the F.T. is simply identical to the L.T. but with the substitution of $s=j\omega$.