Your equations are equivalent to the $1$-form equations
$$
\mathrm{d}f = X_0(f,g)\,\mathrm{d}s + Y_0(f,g)\,\mathrm{d}t 
\quad \text{and}\quad  
\mathrm{d}g = X_1(f,g)\,\mathrm{d}s + Y_1(f,g)\,\mathrm{d}t 
$$
The Frobenius compatibility condition for these two $1$-form equations is indeed the condition that $[X,Y]=0$, where, in the $uv$-plane
$$
X = X_0(u,v)\,\frac{\partial}{\partial u} + X_1(u,v)\,\frac{\partial}{\partial v}\quad \text{and}\quad  
Y = Y_0(u,v)\,\frac{\partial}{\partial u} + Y_1(u,v)\,\frac{\partial}{\partial v}\,.
$$

Note that, setting
$$
X_0(u,v)=e^{-u}\cos(v),\ X_1(u,v)=-e^{-u}\sin(v),\ Y_0(u,v)=e^{-u}\sin(v),\ Y_1(u,v)=-e^{-u}\cos(v),
$$
we get an example (not polynomial, though), for which it turns out that there are maximal solutions (f,g) satisfying your initial conditions for which the range of $(f,g)$ is not all of $\mathbb{R}^2$.  (Maximal solutions are not unique.). The point is that a solution satisfying your initial condition also satisfies
$$
s = e^f\cos g - 1\quad\text{and}\quad t = e^f\sin g,
$$
and these equations are equivalent to the complex equation 
$$
1+ s + i t = e^{f+ig},
$$
so $f+i g$ must be a branch of $\log (1 + s+it)$, which cannot be global.
There is a unique solution on the complement of the ray $t=0$ and $s\le-1$, and if you look at that solution, you will see that $f+ig$ does not map this open set onto the complex plane, in fact, $|g|\le \pi$.