To use the continued fraction for a number to prove its irrationalityit's transcendental, one usually shows that the rational approximations it affords are "too good" for an algebraic number. The traditional tool here is Liouville's theorem, but this has been improved to the more powerful Roth's theorem:
If $\alpha$ is algebraic, then for every $\varepsilon > 0$ there are only finitely many rational numbers $p/q$ satisfying $$ \left|\alpha - \frac{p}{q}\right| < \frac{1}{q^{2+\epsilon}}. $$
Unfortunately, $e$ also satisfies this. Indeed, rational approximations to $e$ are uncommonly bad. As Wadim mentions, there are only finitely many $p/q$ satisfying the following inequality. $$ \left|e - \frac{p}{q}\right| < \frac{\log \log q}{3 q^2 \log q}. $$
But Khinchin proved that, for almost all $\alpha$, $$ \left|\alpha - \frac{p}{q}\right| < \frac{1}{q} \phi(q) $$ has infinitely many solutions if and only if $\sum_q \phi(q)$ diverges.
Additionally, Khinchin showed that the geometric means of the entries of the continued fraction expansion of a real number almost always converge to a universal constant. The geometric means of the entries of the continued fraction expansion of $e$ diverge.
If either of Khinchin's conditions hold for nonquadratic algebraic numbers, the transcendentality of $e$ would follow, but a proof is probably out of reach.
Finally, the continued fraction expansion of $e$ provides an immediate proof of its irrationality, because rational numbers have finite continued fraction expansions. We also have that $e$ is not a quadratic irrational, because those have (eventually) periodic continued fraction expansions.
