Mick commented on my anwer to Is $\lim_{n \to \infty} f^{[n]} ( g^{[n]} (z) ) $ analytic? asking me to look at the related question here. I cannot understand the Op's logic in this question, but there is a simple counter example in the answer I gave to the referenced question.
$$g(z)=z^2+1;\;\;\;b(z)=\sqrt{z};\;\;\;A_n(z)=b^{[\circ n]} \circ g^{[\circ n]}(z);\;\;\;A(z)=\lim_{n\to\infty}A_n(z)$$
(1) is met since $A_n(z)$ It is defined everywhere at the real axis.
(2) is met since the sequence $a_n(z)$ is bounded everywhere in the complex plane. My answer to the referenced question showed that the limit exists and I gave the analytic function for $A(z)$ as a Boetcher function which is defined except at an uncountable number of points corresponding to a Cantor dust. At the Cantor dust points, the limit does not exist because iterating $z \mapsto z^2+1$ we have $\forall n\;|g_n(z)|<2$ so the $g_n$ sequence never grows arbitrarily large. But the $a_n$ sequence is bounded, even though the limit is not defined.
(3) $A(z)$ is strictly increasing at the real axis for positive real(z).
The conjecture says $A(z)$ is defined and converges for: "$z$ with $Re(z) > 0$ And $\Im(z)^2 < 1$". The counter example comes right from my answer for two of the Cantor dust points, $z=0.5\pm i\sqrt{3/4}$. There are an uncountable number of other Cantor dust counter examples which have $|\Im(z)|<1$.
QED
Another perhaps simpler counter-example is iterating $z \mapsto z^2+z\;$ where the Julia interior (where the iteration stays bounded) is a well defined fractal instead of a cantor dust. Here the corresponding $A(z)$ function is also defined and increasing for all positive real values of z, but not at z=0. An example of a counter example on the fractal Julia boundary is z~=0.0762266593+0.922299937i, which is a 3-cycle.
