The only problem is when $\epsilon$ is small, and you do have a singularity at $\epsilon=0$. Now for $x>1,0<\epsilon<1/4$, $$ \frac{x-1}{2}\ln(1-\epsilon)\le\frac{x(-\epsilon/2)}{2} $$ so you can take $a=1/4$. The bounded values of $x$ are unimportant.

I have made an explicit computation and you can find as well an explicit solution with my formula above by plugging the values of $v$ in terms of your $u$. The regularity business, say for the function $v$ follows from the explicit integral expression: you get easily that the $L^2$ norm of $\mathcal K v$ controls the $H^1$ norm in the $x$ variable of $v$. The expression of $w$ shows that you control 2/3 of derivatives for the $z$ variable: if you want an isotropic control then you cannot do better than $2/3$. To see that is not completely obvious: just compute exactly the integral in the phase

@timur It is certainly possible to do what you propose, but you have to keep in mind Hadamard's celebrated sentence about polynomials, in which he said essentially the following: "I do not care so much about approximating the data by polynomials, what matters is how this approximation is transferred to the solution." I will add other comments in a new edit of my answer above.

It seems to me that the formulas above give an explicit integral solution to your problem. From this explicit expression, it is for instance easy to prove hypoellipticity and regularization properties, even to find the exact fractional amount of $z$ derivative that you gain.

@Matt Jacobs I would recommend the Bahouri-Chemin-Danchin book Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 343. Springer, Heidelberg, 2011.

@Matt Jacobs As said in the previous comment, $f(D)u$ is the function whose Fourier transform is $f(\xi)\hat u(\xi)$. The operator $f(D)$ is called a Fourier multiplier for this reason. An integral representation is $$ (f(D)u)(x)=\int e^{2i\pi x\cdot \xi} f(\xi) \hat u(\xi) d\xi, $$ with $$ (\hat u)(\xi)=\int e^{-2i\pi x\cdot \xi} u(x) dx. $$