Uniqueness of exponential function To my knowledge, the exponential function is the unique function satisfying

$f'=f$ and
$f(0)=1$

however, unless I've made a mistake, we have

$$\frac{\partial}{\partial x} (ax)^x = x (ax)^{x-1} a = ax (ax)^{x-1} = (ax)^x$$

and 

$$(a0)^0 = 0^0 =1$$

so I feel like I must be missing something special about $e^x$.  Any pointers would be greatly appreciated.
You differentiated $x^x$ wrong. In fact,

$$  (x^x)' = (e^{x \log x})' \overset{\text{chain rule}}{=} [x \log x]' e^{x\log x} = (\log x +1 )x^x$$

The rule $[x^n]' = n x^{n-1}$ only applies when $n$ is a fixed constant.
The simplest of the mistakes in your derivation is $0^0 = 1$. That is incorrect.
$0^0$ is undefined (it is the same as $0/0$).