Here is a brute-force solution. Note that for $z=a-ib$ with $a,b\in{\mathbb R}$, we have
$$
exp(z)=A-iB \ \textrm{with} \ A=e^a\cos(b), \ B=e^a\sin(b) \tag{1}
$$
and
$$
exp(exp(z))=C-iD \ \textrm{with} \ C=e^A\cos(B), \ D=e^A\sin(B) \tag{2}
$$
Also, let
$$
C^{\sharp}=C-a, D^{\sharp}=D-b \tag{3}
$$
Thus, our goal is find $a,b$ such that $C^{\sharp}=D^{\sharp}=0$ but $(A,B)\neq(a,b)$. Let
$$
\begin{array}{c}
a_1=1.94,\ a_2=2, {\cal A}=[a_1,a_2], \\
b_1=1.4,\ b_2=1.5, {\cal B}=[b_1,b_2], \\
{\cal R}={\cal A}\times{\cal B}
\end{array}\tag{4}
$$
We will show that there is a solution $(a,b)$ in the rectangle $\cal R$. We start by noting
a few generic properties of a point $(a,b)\in{\cal R}$.
Note that ${\cal B}\subseteq (0,\frac{\pi}{2})$, an interval on which $\cos$ is positive and decreasing, and $\sin$ is positive and increasing.
We deduce
$$
\left\lbrace\begin{array}{l}
e^{a_1}\cos(b_2) \leq A \leq e^{a_2}\cos(b_1), \\
e^{a_1}\sin(b_1) \leq B \leq e^{a_2}\sin(b_2)
\end{array}\right.\tag{5}
$$
We can apply this reasoning a second time : the interval in which $B$ lives,
${\cal B}'=[e^{a_1}\sin(b_1),e^{a_2}\sin(b_2)]=[6.85...,7.37...]$ is included in $2\pi+(0,\frac{\pi}{2})$, where again $\cos$ and $\sin$ are positive and monotone. So
$$
\left\lbrace\begin{array}{l}
\cos(e^{a_2}\sin(b_2)) \leq \cos(B) \leq \cos(e^{a_1}\sin(b_1)), \\
\sin(e^{a_1}\sin(b_1)) \leq \sin(B) \leq \sin(e^{a_2}\sin(b_2))
\end{array}\right.\tag{6}
$$
We can now show :
Lemma 1. Fix $a\in {\cal A}$ and let $b$ vary in $\cal B$. Then $C^{\sharp}$ and $D^{\sharp}$ are decreasing in $b$.
Proof of lemma 1. We have $D^{\sharp}=e^{A}\sin(B)-b$ and hence (the differentation denoted by $'$ is with respect to $b$)
$$
(D^{\sharp})'=A'e^{A}\sin(B)+e^{A}B'\cos(B)-1=e^{A}(A\cos(B)-B\sin(B))-1 \tag{7}
$$
where in the last equality we have used $A'=-B$ and $B'=A$. So to show that $D^{\sharp}$ is decreasing, it will suffice to show that the
quantity $q=A\cos(B)-B\sin(B)$ is negative ; but
$$
q \leq A-B\sin(B) \leq e^{a_2}\cos(b_1)-e^{a_1}\sin(b_1)\sin(e^{a_1}\sin(b_1))=-2.46\ldots \tag{8}
$$
For $C^{\sharp}$, the proof is similar and easier : we have
$$
(D^{\sharp})'=A'e^{A}\cos(B)-e^{A}B'\sin(B)=e^{A}(-B\cos(B)-A\sin(B)) \tag{9}
$$
and the negativity of this needs no numerical computation. This finishes the proof of lemma 1.
Lemma 2. Let $a\in {\cal A}$. Then $D^{\sharp}(a,b_1)\gt 0 \gt D^{\sharp}(a,b_2)$.
Proof of lemma 2. We have
$$ D^{\sharp}(a,b_1)=e^{A}\sin(B)-b_1 \geq e^{e^{a_1}\cos(b_1)}\sin(e^{a_1}\sin(b_1))-b_1=0.37\ldots \tag{10}$$.
which shows the fist half of the claim. The proof of the second half is similar. This finishes the proof of lemma 2.
Combining lemma 1 and lemma 2 with the intermediate value theorem, we see that for each $a\in{\cal A}$ there is a unique $\beta(a)\in{\cal B}$ such that $D^{\sharp}(a,\beta(a))=0$. Also, the implicit function theorem tells us that $\beta$ is differentiable, and hence continuous.
Lemma 3. We have $C^{\sharp}(a_1,\beta(a_1))\gt 0 \gt C^{\sharp}(a_2,\beta(a_2))$.
Proof of lemma 3. By numerical computation, we have
$$
D^{\sharp}(a_1,1.44)= -0.07\ldots \tag{11}
$$
Since $D^{\sharp}$ is decreasing in $b$, we deduce $\beta(a_1)\lt 1.44$. But $C^{\sharp}$ is also decreasing in $b$, so that
$$
C^{\sharp}(a_1,\beta(a_1)) \gt C^{\sharp}(a_1,1.44) = 0.08\ldots \tag{12}
$$
Another numerical computation yields
$$
C^{\sharp}(a_2,b_1)= -0.09\ldots \tag{13}
$$
Using again the fact that $C^{\sharp}$ is also decreasing in $b$, we deduce
$$
C^{\sharp}(a_2,\beta(a_2)) \lt C^{\sharp}(a_2,b_1) \lt 0. \tag{14}
$$
This finishes the proof of lemma 3.
Now, the map $a \mapsto C^{\sharp}(a,\beta(a))$ is continuous since $\beta$ and $C^{\sharp}$ are,
so it must have a zero. This gives us a point of periodicity $1$ or $2$, and it cannot be a fixed point because for example
$$
B \gt e^{a_1}\sin(b_1) = 6.85\ldots \gt 1.5 \geq b \tag{15}
$$
so that $B\neq b$. This finishes the answer.
