The standard 1D continuous convolution integral is defined as:
$$y(t) = h(t)*x(t) = \int^{+\infty}_{-\infty}h(\tau)\cdot x(t-\tau)\ d\tau$$
Using fourier transform, $$Y(j\omega) = X(j\omega)\cdot H(j\omega)$$
Therefore, $$X(j\omega) = \frac{Y(j\omega)}{H(j\omega)}$$
Using inverse fourier transform, $$x(t) = \frac{1}{2\pi}\int^{+\infty}_{-\infty}X(j\omega)\cdot e^{j\omega t}\:d\omega$$
Plugging in previous results: $$x(t) = \frac{1}{2\pi}\int^{+\infty}_{-\infty}\frac{Y(j\omega)}{H(j\omega)}\cdot e^{j\omega t}\:d\omega$$
Replacing with fourier transform identities: $$x(t) = \frac{1}{2\pi}\int^{+\infty}_{-\infty}\frac{\int^{+\infty}_{-\infty}y(t')\cdot e^{-j\omega t'}dt'}{\int^{+\infty}_{-\infty}h(t'')\cdot e^{-j\omega t''}dt''}\cdot e^{j\omega t}\:d\omega$$
Can the result be simplified any further, or is the working complete?