Since $e^{i\pi}=-1$, we can rewrite the equation $e^z=-1$ as $e^z=e^{i\pi}$, or equivalently $e^{z-i\pi}=1$.
The solutions of $e^w=1$ are $w=i(2n\pi)$, where $n$ ranges over the integers. Thus the solutions of $e^{z-i\pi}=1$ are $\:i\pi(2n+1)$, where $n$ ranges over the integers.
About your Calculation: Although that calculation happens to give the right answer, the logic is not right. In the calculation, $\sin$ and $\cos$ are functions of a complex variable, and take on complex values.
You are treating the complex $\cos$ and $\sin$ functions as if they had the same formal properties as the corresponding real functions. For example, from $\cos(−iz)+i\sin(−iz)=−1$, you conclude that one of the summands is $0$ and the other is $−1$. But (until one proves otherwise) the imaginary parts of $\cos(−iz)$ and $i\sin(−iz)$ could each be non-zero, but cancel. And as you point out, there is the possibility of considering $\sin(-iz)=i$, $\cos(-iz)=0$. Unfortunately, these are by no means the only possibilities to consider.
Detailed analysis of the complex sine and cosine may enable you to push an argument through. But it is certainly not immediate.