You have put your finger precisely on the statement that is incorrect.
There are two competing conventions with regard to rational exponents.
The first convention is to define the symbol $a^x$ for $a > 0$ only. The symbol $\sqrt[n]{a}$ is defined for negative values of $a$ so long as $n$ is odd, but according to this convention, one wouldn't write $a^{1/n}$, for instance.
In defining $a^{p/q}$ to be $(\sqrt[q]{a})^p$, the author you quoted chose the fraction $p/q$ to be in lowest form so that the definition would be unambiguous. For example, $a^{10/15}$ is defined to be $(\sqrt[3]{a})^2$. However, it is preferable to define $a^{p/q}$ to be $(\sqrt[q]{a})^p$ in all cases and to prove that this definition is independent of the particular representation chosen for $p/q$; this is what more rigorous books tend to do. That is, you prove that if $p/q = r/s$, then $(\sqrt[q]{a})^p = (\sqrt[s]{a})^r$. There is no mention of lowest form.
The competing convention is to also allow $a^x$ to be defined for all $a \ne 0$ and all rational numbers $x = p/q$ that have at least one representation with an odd denominator. You then prove that $(\sqrt[q]{a})^p$ is independent of the particular representation $p/q$ chosen, so long as the denominator is odd. Thus you can write $a^{3/5} = (\sqrt[5]{a})^3 = (\sqrt[15]{a})^{9} = a^{9/15}$. All of that is fine. However, you cannot write $a^{6/10} = (\sqrt[10]{a})^6$, or even $a^{6/10} = \sqrt[10]{a^6}$. The number $a^{6/10}$ is well-defined, but to write down its definition, you must first select a fraction equivalent to $6/10$ that has an odd denominator, which could be $3/5$ or $9/15$ or something else. For $a^{1/2}$, this can't be done at all, so $a^{1/2}$ is undefined for $a < 0$.
The rules for exponents break down if you start allowing $a < 0$ and exponents that can't be written with an odd denominator. For example, the rule $a^{xy} = (a^x)^y$ is valid, but only so long as $x$ and $y$ are both rational numbers that can be written with an odd denominator. This is not the case if you write $a^1 = (a^2)^{1/2}$, despite the fact that both sides of the equation are defined since $a^2 > 0$.
Edit Reading the paper by Tirosh and Even, I was surprised to learn this matter has drawn serious attention from math educators.
A long time ago, I assumed that, apart from complex extensions, $a^x$ for non-integer $x$ should be defined only for $a > 0$. I reasoned that it made no sense to have a function $(-2)^x$ defined only for rational numbers $x$ with odd denominator. I objected strenuously to notations like $(-8)^{1/3}$.
But that was before I taught a calculus class, which is when I realized why some textbook authors are so happy to define $a^x$ for $a < 0$, following the second convention. The reason is that the formula $\frac{d}{dx}(x^r) = rx^{r-1}$ is perfectly valid for $x < 0$ and $r$ with odd denominator.
Under reasonable conditions, at any given moment, only a tiny fraction of water molecules are dissociated into the ${\rm H}^+$ and ${\rm OH}^{-}$ ions. Therefore, the molar concentrations of these ions are very small numbers, so their logarithms are negative. The customary definition of pH takes this into account.
If we add acid to water, then ${\rm H}^+$ ions' concentration grows by orders of magnitude, but still there are way fewer ${\rm H}^+$ ions than water molecules, and the concentration of ${\rm H}^+$ ions still remains a small number. So the logarithm of ${\rm H}^+$ ions' concentration remains negative in the vast majority of situations, and the pH definition still works as intended, stripping the minus sign of the logarithm.
Best Answer
Yes, a complete proof will cover all the cases. But you can simplify the work a lot: note that proving $a^{m+n}=a^m\cdot a^n~~~(\dagger)$ for integers $m,n$ imply the two other properties:
$$\frac{a^m}{a^n}=\begin{cases}a^m\cdot\dfrac 1{a^n}=a^m\cdot a^{-n}\overset{\dagger}{=}a^{m+(-n)}=a^{m-n}&,n\gt 0\\ \dfrac{a^m}{a^0}=a^m=a^m\cdot a^0\overset{\dagger}{=}a^{m-0}&,n=0\\ \dfrac{a^m}{a^{-(-n)}}=a^m\cdot\dfrac 1{a^{-(-n)}}=a^m\cdot a^{-n}\overset{\dagger}{=}a^{m-n}&,n\lt 0\end{cases}$$
$$(a^m)^0:=1=a^{0}=a^{m\cdot 0}$$
$$(a^m)^n=\begin{cases}\underbrace{a^m\cdot a^m\cdots a^m}_{n\text{ times}}\overset{\dagger}{=}a^{\underbrace{m+m+\cdots+m}_{n\text{ times}}}=a^{mn}&,n\gt 0\\ \dfrac 1{\underbrace{a^m\cdot a^m\cdots a^m}_{-n\text{ times}}}\overset{\dagger}{=}\dfrac 1{a^{\underbrace{m+m+\cdots+m}_{-n\text{ times}}}}=\dfrac 1{a^{m(-n)}}=\dfrac 1{a^{-(mn)}}=a^{mn}&,n\lt 0\end{cases}$$
So, all you need to do is to prove $(\dagger)$ for all integers $m,n$
Once you have proved it for positive integers $m,n$, you can wlog consider only one of $m,n$ to be negative and by symmetry, you prove the cases where exactly one of $m,n$ is negative. For the case of both $m,n$ negative, we note that $$a^{m+n}=a^{-(-m-n)}=\frac 1{a^{(-m)+(-n)}}\overset{\dagger}{=}\frac 1{a^{-m}\cdot a^{-n}}=\frac 1{a^{-m}}\cdot\frac 1{a^{-n}}=a^m\cdot a^n$$
where we apply $(\dagger)$ with positive integers $-m,-n$
which leaves us with the case of one or both of $m,n$ be $0$ which is easy to finish up. By symmetry, wlog prove with just one of $m,n$ being $0$ and for the case of $m=n=0$, we have $a^{0+0}=a^{0}=1=1\cdot 1=a^0\cdot a^0$. $_\square$