Solving $e^{e^z}=1$: am I missing something?

Since $1$ can be written $1=e^0$ it follows that a first solution is $$e^z=i2\pi k,\qquad (k \in \mathbb{Z})$$

If $k=0$ there are no solutions since $e^z$ is never zero.

If $k>0$ write $i2 \pi k$ in exponential polar form and you should find that $$|i2 \pi k|=2 \pi |k| = 2 \pi k, \qquad (\text{since $k>0$})$$

So \begin{align} i 2 \pi k &= e^{i \pi /2}e^{\ln(2 \pi k)} \\ &= e^{i \pi /2 + \ln(2 \pi k)} \end{align} Hence solution for $k>0$ is of the form $$z=\ln(2 \pi k)+i\left(\frac{\pi}{2}+2 \pi n \right), \qquad \text{for $n \in \mathbb{Z}, k \in \mathbb{Z} \cap (0, + \infty)$}$$

I leave the case $k < 0$ as an exercise for the OP.

The problem is that your expression "$\ln 2\pi ik$" is multivalued (worse, it is undefined if $k=0$).

One may write $$e^{e^z}=1 \iff e^z = 2\pi i k \stackrel{k\neq 0}{\iff} z = \begin{cases} \ln 2\pi k + \frac{\pi i}{2}(4n+1), & k>0\\ \ln -2\pi k + \frac{\pi i}{2}(4n-1), & k<0 \end{cases}$$

So the solutions are $$z_{n,k} = \ln2\pi |k| + \frac{\pi i}{2}\left(4n+\frac{k}{|k|}\right)$$ for integral $n$ and nonzero integral $k$, and "$\ln$" is the real-valued function of a positive real variable (note that since $k$ is nonzero, one has that $\frac{k}{|k|} = \pm 1$ according to the sign of $k$).