It is true that $\Omega$ looks as an operator of order $0$, sending $L^p$ to $L^p$ ($1<p<\infty$) and $C^\gamma$ into itself ($0<\gamma<1$). Of course when you make a non-linear change of variables in a convolution operator, you loose the translation invariance and $\Omega$ is measuring that distortion.

Yes $\Omega$ is the operator defined in your comment.

You are right, I have changed my answer and added some explanations on Calderon-Zygmund operators.

With the bi-Lipschitz continuity hypothesis, for every $\delta>0$, you get $\epsilon,\sigma >0$ such that $$ \\{\vert G(v)-G(w)\vert\le \sigma\\}\subset\\{\vert v-w\vert\le \delta\\}\subset\\{\vert G(v)-G(w)\vert\le \epsilon\\} $$ implying that last step.

Bazin (mathoverflow.net/users/21907), Motivation for and history of pseudo-differential operators, mathoverflow.net/questions/97604 (version: 2012-05-23)