Is $ds^2=\frac{dxdy}{xy(x-1)(y-1)}$ a valid semi-Riemannian metric?
I got it from adding together four semi-Riemannian metrics:
$ds^2=\frac{dxdy}{xy}+\frac{dxdy}{x-xy}+\frac{dxdy}{y-yx}+\frac{dxdy}{(x-1)(y-1)}=\frac{dxdy}{xy(x-1)(y-1)}$
I have checked that these four metrics are indeed valid semi-Riemannian metrics describing Minkowski space in a unique coordinate system. For example, the first metric in the sum, can be verified to be Minkowski upon inspection. One way to see this is to do a change of coordinates: $u=\frac{dx}{x}$ and $v=\frac{dy}{y}.$
I have verified that the metric $ds^2$ is undefined at $x,y=0,1.$
Your tensor $g$ at a point $(x_0,y_0)$ with $x_0,y_0 \neq 0,1$ is given by $$g = \frac{1}{x_0y_0(x_0-1)(y_0-1)}\mathrm{d}x\mathrm{d}y$$
On each connected component of $\mathbb{R}^{2} \setminus\{(x,y) ~|~ x=0 \text{ or } x=1 \text{ or } y=0 \text{ or } y=1\}$, $g$ is conformal to the tensor $\bar{g} = \pm\mathrm{d}x\mathrm{d}y$, which is symmetric and non-degenerate, of type $(1,1)$. The $\pm$ is because the function $f(x_0,y_0) = \frac{1}{x_0y_0(x_0-1)(y_0-1)}$ is not of constant sign, but it is on every connected component.
Of course the natural coordinates are not globally adapted to the tensor everywhere because of the change of the sign. But the signature of $\pm \mathrm{dx}\mathrm{d}y$ is unvariant and equal to $(1,1)$ so the result follows.
I would add that it seems weird to me to add "the same tensor written in different coordinates system but in the same coordinate system". I mean it has no geometric sense. For example, take the tensor $g=\mathrm{d}x \mathrm{d}y$ on $\mathbb{R}^2$ with natural coordinates. If one takes the $(-x,y)$ coordinates, then $g$ is then written $-\mathrm{d}x \mathrm{d}y$. But the tensor $h = \mathrm{d}x\mathrm{d}y + (-\mathrm{d}x\mathrm{d}y)$ has nothing to do with $"g+g"$ even if $g$ can be written as one or the other in some particular coordinates system.