As defined, for $x$ irrational and $a\in\mathbb{R}_+^*$, there are $A_1 = \left\{r\in\mathbb{Q} | r<x\right\}; A_2 = \left\{s\in\mathbb{Q} | s>x\right\} \Rightarrow B_1 = \left\{a^r|r\in A_1\right\}; B_2 = \left\{a^s|s\in A_2\right\}$. Basically, since $s - r \to 0$, we have:
If $a > 1$, then $r < x < s \Rightarrow a^r < a^x < a^s$;
If $0 < a < 1$, then $r < x < s \Rightarrow a^s < a^x < a^r$;
defining $a^x$. Finally, by this definition, if $A < P < B$ and $A < Q < B$, then we have $P = Q$ (same boundaries, having $B - A \to 0$).
To prove $a^{x + y}$ rule is simple.
If $a > 1 \Rightarrow r < x < s \Rightarrow a^r < a^x < a^s; p < y < q\Rightarrow a^p < a^y < a^q$. By definition these stuff are positive, so: $a^ra^p < a^x a^y<a^sa^q \Leftrightarrow a^{r+p}<a^xa^y<a^{s+q}$. In other hand, $r < x < s$ and $p < y < q \Rightarrow r + p < x + y < s + q \Rightarrow a^{r + p} < a^{x + y} < a^{s + q}$. Both of them have same boudaries, so $a^{x + y} = a^xa^y$. Something similar goes for $0 < a < 1$.
So $1 := a^0 = a^{x +(- x)} = a^{x}a^{-x} \therefore a^{-x} := \dfrac{1}{a^x}$. And there goes $a^{x - y}$ rule: $a^{x - y} = a^{x + (-y)} = a^{x}a^{-y} = a^{x}\dfrac{1}{a^y} = \dfrac{a^x}{a^y}$.
Now suppose $a,b > 1$. Then, $r < x < s \Rightarrow a^r < a^x < a^s; b^r < b^x < b^s \Rightarrow a^rb^r < a^xb^x < a^sb^s \Leftrightarrow (ab)^r < a^xb^x < (ab)^s$. In other hand, $c = ab > 1$, so $c^r < c^x < c^s \Leftrightarrow (ab)^r < (ab)^x < (ab)^s$. Hence $(ab)^x = a^xb^x$. Something similar happens to $0 < a,b < 1$.
The whole problem is when we try to show $(ab)^x = a^xb^x$ for $a > 1$ and $0 < b < 1$. How to show it is also valid in this case?