Let $\sum u_n z^n$ denote the power series of $e^{1/(1-z)}$. As our radius of convergence is $1$, it follows that $u_n$ exhibits sub-exponential growth. On the other hand, $\{u_n\}$ must grow supra-polynomially, else transfer theorems like those found in Singularity Analysis of Generating Functions would then imply that the singularity at $z=1$ is regular. Heuristically, it would appear that $$u_n \sim \alpha n^{-3/4} e^{2\sqrt{n}},$$ for some $\alpha \approx .162982$. This opinion is echoed, without support, as a comment on the OEIS page for A000262. Note: The sequence considered therein is the generating function of $e^{z/(1-z)}$, which has $\mathbb{Z}$ coefficients after scaling $u_n$ by $n!$
How would one derive asymptotic results such as these?
(Edited for spelling.)
Note: I've posted my own solution in the answers below. In short, the saddle-point method applies.
I've managed to answer my own question, but I'll post it here.
As discussed in the comments above, the method of saddle point approximation works here. In short, $f(z):=e^{1/(1-z)}$ meets the sufficient conditions given in Graham et al's Handbook of Combinatorics, II (pg. 1175). It follows that $$[z^n]f(z) \sim (2 \pi b(r_0))^{-1/2} f(r_0) r_0^{-n},$$ where $r_0$ is the saddle point (where $f(z)z^{-n}$ is minimized as a function of $z$) and the function $b(z)$ is given by $$b(z) :=z \left( z \frac{f'(z)}{f(z)}\right)' = \frac{-z(1+z)}{(z-1)^3},$$ the latter only in this instance. Here, our saddle point is $$r_0=\frac{1+2n-\sqrt{1+4n}}{2n},$$ so - putting this all together - we conclude $$ [z^n]f(z) \sim \frac{2^{n-\frac{3}{2}} e^{\frac{2 n}{\sqrt{4 n+1}-1}} \left(\frac{2 n-\sqrt{4 n+1}+1}{n}\right)^{-n}}{\sqrt{\pi } \sqrt{\frac{n \left(4 n^2-3 \sqrt{4 n+1} n+5 n-\sqrt{4 n+1}+1\right)}{\left(\sqrt{4 n+1}-1\right)^3}}} \sim \frac{e^{\sqrt{n}} \left(1-\frac{1}{\sqrt{n}}\right)^{-n}}{2 \sqrt{\pi } \sqrt{n^{3/2}}} \sim \frac{\sqrt{e}}{2\sqrt{\pi}} e^{2 \sqrt{n}} n^{-3/4}.$$ Here, the omitted steps between steps (1) and (2) are easy to do by hand. As for this final step, it suffices to show $$\lim_{n \to \infty} \left(1-\frac{1}{\sqrt{n}}\right)^{-n} e^{-\sqrt{n}}= \sqrt{e},$$ which I'll leave as an interesting exercise. To finish up, we just note that $f(z) = e u(z)$, which gives us asymptotics for $\{u_n\}$. Note: the initial estimate $\alpha \approx .162982$ is quite bad. The actual value is $\approx 0.171099$.