So I can understand your claim, take equation (5) in the original post and substitute ##\gamma^2 =1## instead of the Lorentz factor. This makes the equation become
$$
\big( - \frac{v^2}{c^2} \big)\, (x^2 - c^2\,t^2)
+ 2\,(1-1)\, vtx
= 0\,.
$$
For me to get zero on the left side, I...