Solving base e equation $e^x - e^{-x} = 0$
If $$x = -x$$
Then we can say that $$x = 0$$
Because
$$ x + x = -x + x $$
$$ 2x = 0 $$
$$ x = 0 $$
You are solving for $x$ when $e^x=e^{-x}$. It is not the case for all $x$ that $e^x=e^{-x}$; therefore, the statement $x=-x$ will not be true for all $x$, just the $x$ you are solving for.
Both are correct. $x=-x\Leftrightarrow 2x=0\Leftrightarrow x=0$.
When you take logs of both sides, you don't get $x=-x$. You get $x = -x +2\pi ik$ for integral $k$. This means $2x = 2\pi ik$, so $x = \pi ik$.
The only real-valued solution, then, is $x=0$ (by taking $k=0$).