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 $$ \begin{aligned} X_0(u,v)&=\phantom{-}e^{-u}\cos(v),&& Y_0(u,v)=e^{-u}\sin(v),\\ X_1(u,v)&=-e^{-u}\sin(v),&& Y_1(u,v)=e^{-u}\cos(v), \end{aligned} $$ 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, $|g|\le \pi$.