An approximation by Stirling's formulaLet 0 < α < 1 be a real number and...

Step 1
You are only having difficulty because the version of Stirling you're using is too precise. Instead, use $n!=(1+o(1))\sqrt{2\pi n}(\frac{n}{e}{)}^n$ - replacing all of your $e^{r(n)}$ factors by $1+o(1)$.
Step 2
The step you're stuck on, where you're trying to show that $e^{r(n)-r(\alpha n)-r((1-\alpha )n)}=1+o(1)$, becomes $\frac{1+o(1)}{(1+o(1))(1+o(1))}=1+o(1)$, which is straightforward.
(Technically, the $1+o(1)$ factor we get from Stirling's approximation is $1+o_{n\to \mathrm{\infty }}(1)$, so we also have to observe that $1+o_{\alpha n\to \mathrm{\infty }}(1)$ and $1+o_{(1-\alpha )n\to \mathrm{\infty }}(1)$ are the same asymptotically. This is fine as long as $\alpha$ is a constant, or even more generally than that.)

Step 1
Since $f(n)=o(1)$ just means $\frac{f(n)}{1}\to 0$ as $n\to \mathrm{\infty }$ (i.e. $f(n)=1+o(1)$ means $f(n)\to 1$), our goal is to show that $e^{r(n)-r(\alpha n)-r((1-\alpha )n)}\to 1$, which can be done by showing $r(n)-r(\alpha n)-r((1-\alpha )n)\to 0$. We can use the bounds on r(n) to show that $r(n)-r(\alpha n)-r((1-\alpha )n)<\frac{1}{12n}-\frac{1}{12\alpha n+1}-\frac{1}{12(1-\alpha )n+1}\underset{n\to \mathrm{\infty }}{\to }0$ and $r(n)-r(\alpha n)-r((1-\alpha )n)>\frac{1}{12n+1}-\frac{1}{12\alpha n}-\frac{1}{12(1-\alpha )n}\underset{n\to \mathrm{\infty }}{\to }0$
Step 2
So since $r(n)-r(\alpha n)-r((1-\alpha )n)$ is squeezed between 0 on both sides asymptotically, indeed we have $r(n)-r(\alpha n)-r((1-\alpha )n)\to 0$, and we are done.