The result you state holds because $\mathbb{R}^d$ is simply connected. And (1) and (2) are both true, because you can verify in each case that the right hand side satisfies the properties of the distinguished logarithm on the left hand side, which is uniquely determined by those properties.
Nevertheless, this notation has a potentially confusing issue which you have already pointed out, namely:
$\ln \varphi$ is not to be confused with the composition of a function called "ln" with $\varphi$.
So there is no function $\ln$, only a function $\ln \varphi$. This means that in (2), when you write $\ln \varphi(x) = ix$, the left side must be read as $(\ln \varphi)(x)$ and not as $\ln (\varphi(x))$. If you read it as the latter, then from $\ln(\varphi(x)) = ix$, you would get an apparent contradiction by substituting $x=0$ and then $x = 2 \pi$, since $\varphi(0) = \varphi(2 \pi)$ but $0 \ne 2 \pi i $.
For your last question, the distinguished logarithm does select $ix$ in (2), but it is not as magical as it might initially seem. From the definition of the distinguished logarithm, it is a function $\psi(x)$ satisfying $e^{\psi(x)} = \varphi(x) = e^{ix}$. Of course $\psi(x) = ix$ satisfies that.
Best Answer
Non-zero complex number can be written as: $$z = e^{x + iy} \text{ , where } x,y\in\mathbb R^2$$ From this it follows that $$\log z = x + \log e^{iy}$$
Since $e^{iy}$ is periodic in $y$, you don't have an inverse function. However you can gice a set of possible inverse values much like when you solve equations like $\sin x = 0 \iff x =k\pi$. If you consider the equation:
$$e^{i\varphi} = e^{iy} \iff \varphi = y + k \cdot 2\pi$$
Sometimes it can be useful to use an inverse function, then you can do the same thing as with $sin$ and $cos$. You can restrict the codomain for one period only. The most useful cases would usually be $[-\pi, \pi]$ or $[0, 2\pi]$. In the latter case: $$\log z = x + \log e^{iy} = x + \left\{\frac{y}{2\pi}\right\}\cdot2\pi$$ , where $\{.\}$ is the fractional part function.