If I understand it correctly, [this paper of Abhyankhar][2] says that what you want is not achievable.

Here is what Abhyankar shows. I warn you that I haven't read the paper, and am just skimming the introduction. Let $K$ be the algebraic closure of $k(X,Y)$. So an element $z$ of $K$ should be thought of as a root of a polynomial $z^n + a_{n-1}(X,Y) z^n + \cdots + a_{0}(X,Y)$ whose coefficients depend on two parameters $x$ and $y$. 

Let $L^1$ be the subfield of $K$ generated by all solutions to polynomials of the form $z^n + a_{n-1}(T) z^{n-1} + \cdots + a_{0}(T)$ where $T \in k(X,Y)$ and the $a_i$ are polynomials. This is pretty close to the class of $z$'s you refer to in your update, except that you ask for $T$ and the $a_i$ to be smooth and Abhyankar takes them to be rational functions. Let $L^2$ be the subfield of $K$ generated by solutions to polynomials of the form $z^n + a_{n-1}(T) z^{n-1} + \cdots + a_{0}(T)$ where $T \in L^1$ and the $a_i$ are polynomials. Inductively, make $L^3$, $L^4$, etcetera. Abhyankar shows that $\bigcup L^i$ is strictly smaller than $K$.

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In fact, the meat of Abyhankar's argument is about formal power series, not polynomials. Here I had a little trouble following Abyhankar, so I hope I am summarizing him correctly. Let $\hat{A}$ be the ring of formal power series $k[[X,Y]]$ and let $\hat{K} =\mathrm{Frac} \ \hat{A}$. Let $\hat{L}^1$ be the field generated by all roots of polynomials $\sum a_i(T) z^i$ where $a_i$ are formal power series and $T$ is in the maximal ideal of $\hat{A}$. Let $\hat{B}^1$ be the integral closure of $\hat{A}$ in $\hat{L}^1$. Let $\hat{L}^2$ be the field generated by all roots of polynomials  $\sum a_i(T) z^i$ where $a_i$ are formal power series and $T$ is in the maximal ideal of $\hat{B}^1$. Let $\hat{B}^2$ be the integral closure of $\hat{A}^1$ in $\hat{L}^2$. In this manner, define $\hat{L}^j$ for all $j$. Then $\bigcup \hat{L}^j$ is smaller than the algebraic closure of $\hat{K}$. 

The essence of the proof is that (1) $\hat{L}^{j+1}$ is solvable<sup>1</sup> over $\hat{L}^j$ and (2) the splitting field of $z^6+Xz+Y$ is not solvable over $\mathrm{Frac} \ k[[X,Y]]$.` If you've never thought about these issues before, part (1) may surprise you. To get some intuition for this, note that the root of $x^2 - (1+t)$ is degree $2$ over $\mathrm{Frac} \ k[t]$ but this polynomial factors as $\left( x - \sum \binom{1/2}{j} t^j \right)\left( x + \sum \binom{1/2}{j} t^j \right)$ over $\mathrm{Frac} \ k[[t]]$. So passing to power series can make Galois groups smaller.

You might enjoy thinking about the topology in part (2). Here is as far as I have gotten. The polynomial $z^6+Xz+Y$ has a multiple root if and only if $5^5 X^6 + 6^6 Y^5=0$. This is a real $2$-fold in $\mathbb{C}^2$, with a singularity at $(0,0)$. Its intersection with a sphere around the origin is a torus knot of type $(6,5)$. The roots of $z^6+Yz+X$ form a $6$-sheeted cover of this torus-knot complement. Abyhankar's claim is that the monodromy of this cover is not solvable.
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So, Abhyankar is disproving your hope not just for polynomials but for formal power series. The map "take the Taylor series" is a map from the ring of smooth functions to the ring of formal power series. (This map has a kernel, containing things like $e^{-1/x^2}$.) I think (but have not checked in detail) that this map should mean that Abhyankar's result also works for smooth functions.

<sup>1</sup> I think that, at $j=0$, this statement is only right if we take $k$ algebraically closed, although I don't see where Abhyankar says this. But I'm perfectly willing to do that. Note that $\hat{L}^1$ contains the algebraic closure of $k$, so the issue of whether or not $k$ is algebraically closed goes away once we get one step up the tower. Also, I think that when $k=\mathbb{C}$, we should be able to strengthen solvable to abelian.

  [1]: http://en.wikipedia.org/wiki/Hilbert%2527s_thirteenth_proble
  [2]: http://www.emis.de/journals/SC/1997/2/pdf/smf_sem-cong_2_1-11.pdf