In the complex numbers, unlike in the case of real numbers, the function $e^z$ is not one-to-one. Therefore it does not have an inverse that we can call "log". Essentially, the difficulty arises from the fact that $e^{2 \pi i n} = 1$ for every integer $n$. Thus, "$\log 1$" could be any of the numbers $2 \pi i n$. In general, there will be infinitely many possible choices for the log of any nonzero complex number, all of which differ from each other by an integer multiple of $2 \pi i$. So when we talk about "$\log z$" over the complex numbers, we have to specify which choice we are making for each value of $z$.
To define $a^z$, you just need to choose a value of $\log{a}$, and then define $a^z = e^{z \log{a}}$, as you said. Given a nonzero value of $a$, there are many choices for $\log{a}$, but once you fix it, then $a^z$ is a perfectly well-defined analytic function for all $z$ (since you can use the same choice of $\log{a}$ for all $z$.) No principal branches involved.
There is a problem, however, with defining $z^a$. (Is this what you meant?) Now you have $z^a = e^{a \log{z}}$, and you have to make a choice of $\log{z}$ for each nonzero $z$. Making a random choice for each nonzero $z$ would produce a badly behaved function; the interesting question is whether we can make the choices so that the resulting function is continuous for all nonzero $z$.
It turns out, unfortunately, that we cannot make a continuous choice of $\log{z}$ for all $z \ne 0$. The main difficulty can be described as follows. As you probably know, every $z \ne 0$ can be written in the polar form $z = r e^{i \theta}$ for real numbers $r, \theta$, and $r>0$. Geometrically, $r$ is the distance of $z$ from the origin, and $\theta$ is the angle from the positive $x$-axis to vector from 0 to $z$. Thus we can "define" $\log{z}$ by $$"\log{z} = (\log{r}) + i \theta "$$
where I have put the equation in quotes because we have to be careful how we interpret it.
First of all, what is $\log{r}$? This is no problem: $r$ is a positive real number, which therefore has a unique real logarithm. We define $\log{r}$ to be the real logarithm of $r$. The problem comes with the imaginary part: we have lots of choices for $\theta$, because $0, 2\pi, 4\pi, \ldots, -2\pi, -4\pi, \ldots$ all represent the same angle. So the question boils down to: Can we make a continuous choice of $\theta$ for all complex numbers $z \ne 0$?
We can't. Let's say we choose the value $\theta=0$ at the point $z=1$. Now follow the circle of radius 1 centered at 0, walking counterclockwise along the circle. Now if you think about it carefully, the choice of $\theta$ at each point along this circle is completely determined by the choice of $\theta=0$ at the point $z=1$, if we want $\theta$ to be continuous. Thus, on the line $y=x$ (for positive $x$), we have to choose $\theta=\pi/4$, not something else like $9\pi/4$; on the positive $y$-axis, we have to choose $\theta=\pi/2$, not $5\pi/2$, etc. But now when we walk all the way around the circle, back to the point $z=1$, we find that our original choice 0 is now forced to be $2\pi$. Thus $\theta$ cannot be defined continuously everywhere away from 0.
What can we salvage? We can try to define $\theta$ continuously on a smaller subset of the nonzero complex numbers, instead of all of them. Essentially, the problem was that looping around the origin once increases $\theta$ by $2 \pi$. If we remove enough complex numbers so that it is impossible to loop around the origin without hitting one of these removed points, then $\theta$ can be defined continuously on the remaining points. These removed points are called a "branch cut".
A common choice of branch cut is the negative real numbers. It's not hard to convince yourself that you can't make a loop around the origin without hitting a negative real number at some point. And you can define $\theta$ continuously for all $z$ excluding the real numbers $z \le 0$. If you choose $\theta=0$ for $z=1$, as before, then for points $z$ slightly "above" the negative real axis, we choose $\theta$ positive and having magnitude slightly less than $\pi$; for points $z$ slightly "below" the negative real axis, we choose $\theta$ negative and having magnitude slightly less than $\pi$. All other points fill in naturally. Note that there is a discontinuity as you cross the branch cut (the negative real numbers), as expected.
Once we define $\theta$, we can define $\log z = (\log r) + i \theta$, and thus define $z^a = e^{a \log{z}}$, which was our original goal.
Having said all that, there is one case where all these choices don't matter, and we can define $z^a$ continuously over all nonzero $z$. What case is that? That is the case where $a$ is an integer: For in that case, making a different choice of $\log{z}$ changes the value of $a \log{z}$ by an integer multiple of $2 \pi i$ and thus the value of $e^{a \log{z}}$ isn't affected at all. So the value of $z^a$ is independent of the choice of $\log{z}$ when $a$ is an integer. And if you think about it, there's a good reason for that: for example, when $a=3$, the value of $z^3$ had better not be any number other than $z \cdot z \cdot z$ !
Best Answer
The principal value of the complex logarithm and the principal value of the argument of non-zero complex numbers are connected via $$ \DeclareMathOperator{Log}{Log} \DeclareMathOperator{Arg}{Arg} \Log(z) = \log |z| + i \Arg(z) \, , $$ so the question whether $$ \tag{1} \Log(z_1 z_2) = \Log(z_1) + \Log(z_2) $$ holds or not is equivalent to asking if $$ \tag{2} \Arg(z_1 z_2) = \Arg(z_1) + \Arg(z_2) $$ holds or not.
The number on the right-hand side of $(2)$ is an argument of $z_1 z_2$, and $\Arg(z)$ is usually defined as the argument in the range $(-\pi, \pi]$. This means that $(2)$ and $(1)$ hold if and only if $$ -\pi < \Arg(z_1) + \Arg(z_2) \le \pi \, . $$
You have restricted $z_1$ and $z_2$ to the upper half-plane $\Im(z) > 0$, but that is not sufficient to make $(1)$ and $(2)$ hold. We can still find numbers whose arguments add up to a value larger than $\pi$, e.g. by picking both numbers in the second quadrant.
A simple counterexample is $z_1 = z_2 = -1 + i$.