Here is my attempt to show that it doesn't depend on $g$:
In my opinion the solution lays in the origin of the FDM schemes. The central difference scheme is the result of the subtraction of the Taylor series: $$f(x_{j+1}) = f(x_j)+h\cdot f'(x_j)+h^2\cdot \frac{f''(x_j)}{2}+h^3 \cdot \frac{|f'''(\xi_1)|}{6} , \xi_1 \in[x_i,x_{i+1}]$$
$$f(x_{j-1}) = f(x_j)-h\cdot f'(x_j)+h^2\cdot \frac{f''(x_j)}{2}-h^3 \cdot \frac{|f'''(\xi_2)|}{6} , \xi_2 \in[x_{i-1},x_{i}]$$
$$f(x_{j+1}) - f(x_{j-1}) = 2h \cdot f'(x_j) + \frac {h^3}{6} \cdot (|f'''(\xi_1)| + |f'''(\xi_2)|)$$
If we maximize the error terms we can substitute them with:
$$\frac{|f'''(\xi_1)| + |f'''(\xi_2)|}{6} \leq \frac {max(f'''(\xi))}{3} = C, \xi \in [x_{j-1},x_{j+1}]$$
Which leads us to the final definition of the central difference scheme:
$$f'(x_j)=\frac {f(x_{j+1}) - f(x_{j-1})}{2h}+\tilde{C}h^2$$
Let us take a general case for a differential equation:
$$g(u'(x))=f(x)$$
On our finite domain $\Omega$ with our differential matrix $A$ and right side $F_i$, as well as our local error $R_i$, which is linearly dependent on $h^p$ ($p =$ order of convergence, $h=$ stepsize):
$$g(A \space u_i + R_i) = F_i$$
Supposing $g(x)$ has an inverse and our matrix $A$ is well defined (problem is well posed...):
$$A \space u_i = g^{-1}(F_i) - R_i$$
$$u_i = A^{-1}(g^{-1}(F_i) - R_i)$$
Here you can observe, that the error remains the same as for the case of having $g(x) = ax+b$ (linear). I am not a mathematician so don't quote me here, but another hint to my opinion might give you the relation (Burgers equation):
$$uu_x = \frac{1}{2}u_x^2$$
What do you think?
