Still thinking about the interesting question!

***Not an answer***, but too big for a comment.

To show what I meant in my comment to Daniel Litt's answer about the difference between uniform *absolute* convergence and ordinary uniform convergence:

I **think** (but it was a while ago when I thought about it) that there exist $u_0, u_1, \ldots$ such that $\sum_{n=0}^\infty u_n z^n$ converges uniformly for all $|z| \leq 1$, but not uniformly absolutely, i.e. $\sum_n |u_n| = +\infty$, but also
$$
\forall \\, \epsilon>0, \quad \exists \\, N(\epsilon)>0 \quad \text{such that}: \qquad
\forall \\, |z| \leq 1, \quad \forall \\, m \geq n > N(\epsilon), \qquad
\left| \sum_{k=n}^{m} u_k z^k \right| < \epsilon.
$$
This function $f(z) = \sum_{n=0}^\infty u_n z^n$ satisfies
$$
f \in A(D) \setminus W_+(D),
$$
which shows in particular that the *disc algebra* $A(D)$, of continuous functions on the closed unit disc  which are analytic on the open unit disc, is strictly larger than the *Wiener algebra* $W_+(D)$ of power series absolutely convergent on the closed unit disc.


So the problem is going to be pretty hard because we can't use **absolute** convergence (unless I'm being stupid or there is a clever trick exploiting the special structure of the exponential functions).

The easiest "solution" is just to ignore it (i.e. ***assume*** absolute convergence!) This gives a slightly different, but still interesting, question.