Let's simplify your problem a bit:

You showed that 2ln(-1) = ln(1)

That statement an be written as e^(2i*pi) = e^(0).

Are you sure that we can conclude that 2i*pi = 0?

Let us give an easier example to understand.

1⁵=1⁶ does not mean that 5=6 because functions like 1^(x),0^(x), -1^(x) can have the same values for different values of x (eg -1²=-1⁴=1)

Let us take this specific example,

e^(i*pi*2) = e^(i*pi*0) => (e^(i*pi))^(2)=(e^(i*pi))^(0)

Substituting e^(i*pi)=(-1); We have -1^(2) = -1^(0)

We already know from a previous example that -1^(x) can have same value for different x so the same applies for e^(i*pi*x).

Hence, we can avoid ln(-1) as it can have multiple different answers. Better to avoid it as we do for log base 1, 0.

This may not be the best way to approach this problem nor the most accurate way either. I haven't ever dealt with complex analysis but this is the way I like to see it and I think this is intuitive for a layman or a novice like me.