Is there a trick to establish $z = \tan \left[ \frac{1}{i} \log \left( \sqrt{ \frac{1+iz}{1-iz} } \right) \right]$?

Hint Inverse trigonometric functions are usually extended to the complex plane as follows: $$\arctan z = \int_0^z \frac{\mathrm dx}{1 + x^2},\qquad z \notin \{i,-i\},$$ where the part of the imaginary axis that does not strictly lie between $-i$ and $i$ is the cut between the principal branch and the other branches.

One can then express the inverse tangent via the complex logarithm: $$\arctan z = \frac12i\left[\ln(1 - iz) - \ln(1 + iz)\right].$$

Your idea of writing $$\tan x = \frac{ e^{ix } - e^{-ix} }{i( e^{ix} + e^{-ix} ) },$$ is a good start. Substituting $x=\frac{1}{i}\log\sqrt{\tfrac{1+iz}{1-iz}}$ shows that $$\tan x=\frac{1}{i}\frac{\sqrt{\frac{1+iz}{1-iz}}-\sqrt{\frac{1-iz}{1+iz}}}{\sqrt{\frac{1+iz}{1-iz}}+\sqrt{\frac{1-iz}{1+iz}}}.$$ Mutiplying the numerator and denominator by $\sqrt{(1+iz)(1-iz)}$ yields $$\tan x=\frac{1}{i}\frac{(1+iz)-(1-iz)}{(1+iz)+(1-iz)}=\frac{1}{i}\frac{2iz}{2}=z,$$ which isn't as nasty a calculation as you might have feared.

The following argument establishes a definite domain $\Omega'\subset {\mathbb C}$ on which the stated identity is valid using the principal values of $\sqrt{\mathstrut}$ and $\log$.

We start with the slit region $\Omega:=\{z\in{\mathbb C}\>|\>z\ne it, t\in{\mathbb R}, |t|\geq1\}$. The Moebius map $$T: \quad z\mapsto w:={1+iz\over 1-iz}$$ maps $\Omega$ onto the $w$-plane with the negative real axis removed. The principal value $w\mapsto\sqrt{w}$ is well defined there and maps the slit $w$-plane onto the right half plane. On the latter, the principal value of the logarithm is well defined, so that we arrive at ${\rm Log}\sqrt{w}$, situated in a horizontal strip of width $\pi$ with the real axis as centerline. The points $$\zeta:={1\over i}{\rm Log}\sqrt{w}$$ therefore fill the vertical strip $$S:=\left\{\zeta=\xi+i\eta\in{\mathbb C}\>\biggm|\>-{\pi\over2}<\xi<{\pi\over 2}\right\}\ .$$ Put $\Omega':=\Omega\cap S$. I claim that one has $$q(z):=\tan\left({1\over i}{\rm Log}\sqrt{1+iz\over 1-iz}\right)=z\qquad\forall z\in \Omega'\ .$$ Proof. By definition, $$\tan\zeta={1\over i}{e^{i\zeta}-e^{-i\zeta}\over e^{i\zeta}+e^{-i\zeta}}\ .$$ In the case at hand we have $\zeta:={1\over i}{\rm Log}\sqrt{w}$, so that we obtain $$q(z)={1\over i}{\sqrt{w}-1/\sqrt{w}\over \sqrt{w}+1/\sqrt{w}}={1\over i}{w-1\over w+1}=\ldots=z\ .$$