If I remember correctly, this is impossible for any (nonempty) simply-connected compact manifold of positive dimension. In particular, $S^2$ cannot have such an atlas. Off the top of my head, I don't remember where I saw this statement, but I'll try to find a reference.

Followup: I still don't remember the reference, but I think that the proof goes something like this:

By hypothesis, both $\tau=\psi\circ\phi^{-1}$ and $\sigma = \phi\circ\psi^{-1}$ are polynomial maps, and and hence they are defined everywhere on $\mathbb{R}^n$. By hypothesis there is a nonempty open domain $D_1\subset\mathbb{R}^n$ such that $\sigma(\tau(x)) = x$ for all $x\in D_1$ and a nonempty open domain $D_2\subset\mathbb{R}^n$ such that $\tau(\sigma(y)) = y$ for all $y\in D_2$. Since $\sigma$ and $\tau$ are polynomial mappings, it follows that $\sigma(\tau(x)) = x$ for all $x\in\mathbb{R}^n$ and $\tau(\sigma(y)) = y$ for all $y\in\mathbb{R}^n$. Thus, $\sigma$ and $\tau$ are globally invertible. Moreover, by the chain rule, we have $\sigma'(\tau(x))\tau'(x) = I$ for all $x\in \mathbb{R}^n$, and, taking determinants, it follows that $\det(\tau'(x))$ and $\det(\sigma'(\tau(x)))$, which are polynomial in $x$ must actually be (nonzero) constants. Consequently, it follows that $\Lambda^n(T^*M)$ carries a (unique) flat connection such that, if $\mathrm{d}x$ is the standard volume form on $\mathbb{R}^n$, then $\psi^*(\mathrm{d}x)$ is a (local) parallel section of $\Lambda^n(T^*M)$ for each $(\psi,U)$ in the atlas $\mathcal{A}$. By simple-connectivity, there is a global parallel volume form $\mu$ on $M$, and we can, by scaling the members of our atlas, arrange so that $\psi^*(\mathrm{d}x) = \mu$ on $U$ for all $(\psi,U)$ in the atlas.

Now, it is easy to see, using the global nature of the polynomial transition functions that any $(\psi,U)\in\mathcal{A}$ can be uniquely analytically continued along any smooth path in $M$. Since $M$ is simply connected, this means that $\psi$ can be extended uniquely analytically to all of $M$, and hence $\psi:M\to\mathbb{R}^n$ is a smooth map that satisfies $\psi^*(\mathrm{d}x) = \mu$ on all of $M$. Obviously, this is impossible.

If I remember correctly, this is impossible for any (nonempty) simply-connected compact manifold of positive dimension. In particular, $S^2$ cannot have such an atlas. Off the top of my head, I don't remember where I saw this statement, but I'll try to find a reference.

Followup: I still don't remember the reference, but I think that the proof goes something like this:

By hypothesis, both $\tau=\psi\circ\phi^{-1}$ and $\sigma = \phi\circ\psi^{-1}$ are polynomial maps, and hence they are defined everywhere on $\mathbb{R}^n$. By hypothesis there is a nonempty open domain $D_1\subset\mathbb{R}^n$ such that $\sigma(\tau(x)) = x$ for all $x\in D_1$ and a nonempty open domain $D_2\subset\mathbb{R}^n$ such that $\tau(\sigma(y)) = y$ for all $y\in D_2$. Since $\sigma$ and $\tau$ are polynomial mappings, it follows that $\sigma(\tau(x)) = x$ for all $x\in\mathbb{R}^n$ and $\tau(\sigma(y)) = y$ for all $y\in\mathbb{R}^n$. Thus, $\sigma$ and $\tau$ are globally invertible. Moreover, by the chain rule, we have $\sigma'(\tau(x))\tau'(x) = I$ for all $x\in \mathbb{R}^n$, and, taking determinants, it follows that $\det(\tau'(x))$ and $\det(\sigma'(\tau(x)))$, which are polynomial in $x$ must actually be (nonzero) constants.

Consequently, it follows that $\Lambda^n(T^*M)$ carries a (unique) flat connection such that, if $\mathrm{d}x$ is the standard volume form on $\mathbb{R}^n$, then $\psi^*(\mathrm{d}x)$ is a (local) parallel section of $\Lambda^n(T^*M)$ for each $(\psi,U)$ in the atlas $\mathcal{A}$. By simple-connectivity, there is a global parallel volume form $\mu$ on $M$, and we can, composing the members of our atlas with linear transformations of $\mathbb{R}^n$, get a new atlas with polynomial transitions that satisfies $\psi^*(\mathrm{d}x) = \mu$ on $U$ for all $(\psi,U)$ in the new atlas.

There is also a way to analytically continue the map $\psi$ coming from a chart $(\psi,U)$ to the domain $U\cup V$ for any chart $(\phi,V)$ for which $U\cap V$ is non-empty: Since $\tau$ is defined on all of $\mathbb{R}^n$, and since $\psi = \tau\circ\phi$ on $U\cap V$, one can take $\psi(q) = \tau(\phi(q))$ for all $q\in V$, which extends $\psi$ to $V$.

Using this construction, any $(\psi,U)\in\mathcal{A}$ can be uniquely analytically continued along any smooth path in $M$, and one easily verifies that this extension depends only on the homotopy class of the path with fixed endpoints. Since $M$ is simply connected, this means that $\psi$ can be extended uniquely analytically to all of $M$, and hence $\psi:M\to\mathbb{R}^n$ is a smooth map that satisfies $\psi^*(\mathrm{d}x) = \mu$ on all of $M$. Obviously, this is impossible.

If I remember correctly, this is impossible for any simply-connected compact manifold. In particular, $S^2$ cannot have such an atlas. Off the top of my head, I don't remember where I saw this statement, but I'll try to find a reference.

If I remember correctly, this is impossible for any (nonempty) simply-connected compact manifold of positive dimension. In particular, $S^2$ cannot have such an atlas. Off the top of my head, I don't remember where I saw this statement, but I'll try to find a reference.

Followup: I still don't remember the reference, but I think that the proof goes something like this:

By hypothesis, both $\tau=\psi\circ\phi^{-1}$ and $\sigma = \phi\circ\psi^{-1}$ are polynomial maps, and and hence they are defined everywhere on $\mathbb{R}^n$. By hypothesis there is a nonempty open domain $D_1\subset\mathbb{R}^n$ such that $\sigma(\tau(x)) = x$ for all $x\in D_1$ and a nonempty open domain $D_2\subset\mathbb{R}^n$ such that $\tau(\sigma(y)) = y$ for all $y\in D_2$. Since $\sigma$ and $\tau$ are polynomial mappings, it follows that $\sigma(\tau(x)) = x$ for all $x\in\mathbb{R}^n$ and $\tau(\sigma(y)) = y$ for all $y\in\mathbb{R}^n$. Thus, $\sigma$ and $\tau$ are globally invertible. Moreover, by the chain rule, we have $\sigma'(\tau(x))\tau'(x) = I$ for all $x\in \mathbb{R}^n$, and, taking determinants, it follows that $\det(\tau'(x))$ and $\det(\sigma'(\tau(x)))$, which are polynomial in $x$ must actually be (nonzero) constants. Consequently, it follows that $\Lambda^n(T^*M)$ carries a (unique) flat connection such that, if $\mathrm{d}x$ is the standard volume form on $\mathbb{R}^n$, then $\psi^*(\mathrm{d}x)$ is a (local) parallel section of $\Lambda^n(T^*M)$ for each $(\psi,U)$ in the atlas $\mathcal{A}$. By simple-connectivity, there is a global parallel volume form $\mu$ on $M$, and we can, by scaling the members of our atlas, arrange so that $\psi^*(\mathrm{d}x) = \mu$ on $U$ for all $(\psi,U)$ in the atlas.

Now, it is easy to see, using the global nature of the polynomial transition functions that any $(\psi,U)\in\mathcal{A}$ can be uniquely analytically continued along any smooth path in $M$. Since $M$ is simply connected, this means that $\psi$ can be extended uniquely analytically to all of $M$, and hence $\psi:M\to\mathbb{R}^n$ is a smooth map that satisfies $\psi^*(\mathrm{d}x) = \mu$ on all of $M$. Obviously, this is impossible.