From an 2015 MSE answer called Indefinite double integral
we conclude that the general solution of the PDE
$$
\frac{\partial^2 u}{\partial p\,\partial q} = \frac{\partial^2 u}{\partial q\,\partial p} = 0
$$
is given by
$$
u(p,q) = F(p) + G(q)
$$
The wave equation has been considered as an example:
$$
\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} = 0
$$
With a lemma in the
abovementioned reference, this is converted to:
$$
\frac{\partial}{\partial (x-ct)}\frac{\partial}{\partial (x+ct)} u =
\frac{\partial}{\partial (x+ct)}\frac{\partial}{\partial (x-ct)} u = 0
$$
So we find that the general solution of our wave equation is given by:
$$
u(x,t) = F(p) + G(q) = F(x-ct) + G(x+ct)
$$
Here $\,x-ct=0\,$ and $\,x+ct=0\,$ are the characteristics.
Quite analogously, if the OP's simplified equation $\,u_{xy}=0\,$ is considered instead, then the solution must be:
$$
u(x,y) = F(x) + G(y)
$$
with $\,x=0\,$ and $\,y=0\,$ as the two characteristics. Indeed the equation is still hyperbolic, because $b^2-ac=b^2=1\gt 0$.
Let's try another hyperbolic argument :-)
Both the equations $\,xy=1\,$ and $\,x^2-y^2=2\,$ represent an orthogonal hyperbola, right?
They are equivalent up to a rotation over $45^o$.
Let's try the same for the equation $\,u_{xy}=0\,$.
Consider the rotation transformation
$$
\begin{cases}
x_1 = \cos(\alpha)\,x-\sin(\alpha)\,y \\
y_1 = \sin(\alpha)\,x+\cos(\alpha)\,y
\end{cases}
$$
We want to know how the derivatives of the solution $\,u\,$ transform.
$$
\frac{\partial u}{\partial x} =
\frac{\partial u}{\partial x_1}\frac{\partial x_1}{\partial x}
+ \frac{\partial u}{\partial y_1}\frac{\partial y_1}{\partial x} \\
\frac{\partial u}{\partial y} =
\frac{\partial u}{\partial x_1}\frac{\partial x_1}{\partial y}
+ \frac{\partial u}{\partial y_1}\frac{\partial y_1}{\partial y}
$$
It follows, with operator notation:
$$
\frac{\partial}{\partial x} u =
\left[ \cos(\alpha)\frac{\partial}{\partial x_1}
+\sin(\alpha)\frac{\partial}{\partial y_1} \right] u\\
\frac{\partial}{\partial y} u =
\left[ -\sin(\alpha)\frac{\partial}{\partial x_1}
+\cos(\alpha)\frac{\partial}{\partial y_1} \right] u
$$
So we have:
$$
\frac{\partial^2}{\partial x \, \partial y} =
\left(\frac{\partial}{\partial x}\right)\left(\frac{\partial}{\partial y}\right) = \\
\left[ \cos(\alpha)\frac{\partial}{\partial x_1}
+\sin(\alpha)\frac{\partial}{\partial y_1} \right]
\left[ -\sin(\alpha)\frac{\partial}{\partial x_1}
+\cos(\alpha)\frac{\partial}{\partial y_1} \right] = \\
-\sin(\alpha)\cos(\alpha)\left(\frac{\partial}{\partial x_1}\right)^2
+\left[\cos^2(\alpha)-\sin^2(\alpha)\right]\frac{\partial}{\partial x_1}\frac{\partial}{\partial y_1}
+\sin(\alpha)\cos(\alpha)\left(\frac{\partial}{\partial y_1}\right)^2 = \\
-\frac12 \sin(2\alpha)\frac{\partial^2}{\partial x_1^2}
+ \cos(2\alpha)\frac{\partial^2}{\partial x_1\,\partial y_1}
+ \frac12 \sin(2\alpha)\frac{\partial^2}{\partial y_1^2}
$$
Now choose $\,\alpha = 45^o = \pi/4\,$, just as with the orthogonal hyperbolas, then we have:
$$
\frac{\partial^2}{\partial x \, \partial y} = -\frac12 \frac{\partial^2}{\partial x_1^2} + \frac12 \frac{\partial^2}{\partial y_1^2}
$$
And so:
$$
\frac{\partial^2 u}{\partial x \, \partial y} = - \frac12 \left[ \frac{\partial^2 u}{\partial x_1^2} - \frac{\partial^2 u}{\partial y_1^2}\right] = 0
$$
Which is clearly a hyperbolic, not a parabolic, equation.
The characteristics $\,x=0\,$ and $\,y=0\,$ are likewise rotated, giving $\,x_1+y_1=0\,$ and $\,x_1-y_1=0\,$.