Is this generalization of an exercise in Stein true?

The following question is exercise $14$ in chapter $2$ in Stein and Shakarchi's Complex Analysis.

Suppose that $f$ is holomorphic in an open set containing the closed unit disc, except for a pole at $z_0$ on the unit circle. Show that if $$\sum_{n=0}^\infty a_nz^n$$ denotes the power series expansion $f$ in the open unit disc, then $$\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=z_0.$$

A solution may be found here or here. A discussion in the comments of one of the linked questions prompted me to ask another question.

Question: Does this remain true when 'pole' is replaced by 'essential singularity'?

I strongly suspect the answer is no, but I can't find a convenient counterexample. It seems like something like $e^{\frac{1}{z-1}}$ should provide a counterexample, but the calculus involved in computing the power series coefficients is very messy. I recall there being some standard power series that people always use to show that anything can happen on the boundary of a disk of convergence, but I can't find them at the moment. Perhaps one of those works?

Short version. The function $$g(x,z)=\frac{1}{1-z} \exp\left(-\frac{xz}{1-z}\right) \tag1$$ is the generating function for Laguerre polynomials; that is, $$g(x,z)=\sum_{n=0}^\infty L_n(x)\,z^n \tag2$$ Set $x=1$. The function $g(1,z)$ has an essential singularity at $z=1$ (it's almost the same as the function in the OP). That the ratio $L_n(1)/L_{n+1}(1)$ does not approach $1$ (or any other number) follows from the asymptotic expansion
$$L_n(1)= \frac{\sqrt{e/\pi}}{n^{1/4}} \cos(2\sqrt{n}-\pi/4)+O(n^{-3/4}) \tag3$$ which shows that $L_n(1)$ changes sign infinitely often (though increasingly slowly).

Longer but self-contained version. The function $$g(z)=\frac{1}{1-z} \exp\left(-\frac{z}{1-z}\right) \tag 4$$ satisfies $g'(z)=-z\, g(z)/(1-z)^2$, which can be written as $$g'(z)-2zg'(z)+z(zg(z))'=0 \tag5$$ Plugging $g(z)=\sum_{n=0}^\infty a_n z^n$ in (5) and extracting the coefficient of $z^n$, we find $$(n+1)a_{n+1} - 2n a_n +n a_{n-1} =0,\quad n\ge 1 \tag6 $$

Suppose $a_n$ have constant sign for $n\ge N$; without loss of generality $a_n\ge 0$ for $n\ge N$. By making $N$ larger, we can arrange $a_N>0$. Only $n\ge N$ are considered in the sequel.

Let $\Delta_n=a_{n+1}-a_n$. From (6) we have $$\Delta_{n+1} - \Delta_n = a_{n+1}-(2a_n-a_{n-1}) = -\frac{a_{n+1}}{n+1} \le 0 \tag7$$ Thus, $\Delta_n$ is a nonincreasing sequence (which makes $a_n$ a concave sequence). If some $\Delta_n$ is negative, the sequence $a_n$ will also become negative, contrary to assumption. Thus, $a_n$ is nondecreasing, hence $a_n\ge a_N$. Summing (7) over $n=N,N+1,\dots , M$ we get $$\Delta_{M+1}-\Delta_N \le -a_N \sum_{n=N}^{M}\frac{1}{n+1} \tag8$$ Since the harmonic series diverges, (6) implies $\Delta_{M+1}<0$ for large $M$, a contradiction.

Conclusion: $a_n$ changes sign infinitely often. In particular, $a_n/a_{n+1}\not\to 1$.

Remark: (6) is secretly the recurrence relation for Laguerre polynomials.