Can you prove the power rule for irrational exponents without invoking $e$?

Yes. In general, if you have a sequence of $C^1$ functions $f_n$ on some interval and functions $f$ and $g$ such that $f_n\to f$ pointwise and $f_n'\to g$ uniformly, then $f'=g$. (Quick sketch of how to prove this without using integration: fix $x$ and $h$ and pick $n$ so that $f_n$ is sufficiently close to $f$ at $x$ and $x+h$ and $f_n'$ is sufficiently close to $g$ uniformly. Then $\frac{f(x+h)-f(x)}{h}$ will be close to $\frac{f_n(x+h)-f_n(x)}{h}$. By the mean value theorem the latter is equal to $f_n'(y)$ for some $y$ between $x+h$ and $x$, and $f_n'(y)$ is close to $g(y)$. Finally, if $h$ is sufficiently small, $g(y)$ will be close to $g(x)$ since $g$ is continuous.)

So, given $r\in\mathbb{R}$, pick a sequence of rational numbers $q_n$ converging to $r$ and let $f_n(x)=x^{q_n}$. Then $f_n(x)$ converges to $f(x)=x^r$ and $f_n'(x)=q_nx^{q_n-1}$ converges to $g(x)=rx^{r-1}$. Moreover, it is not hard to check that these convergences are uniform on any compact subset of $(0,\infty)$. It thus follows that $f'=g$ on $(0,\infty)$.