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I think the answer is yes, but there are details I am missing.

Here is my attempt:

But now that I think about it, $g$ doesn't need to be $C^1$. Unless I have missed something, $f(z) = (z-1)^2 e^{1/(z-1)}$, as a function on $\bar{\Delta}$, is differentiable at $z=1$ (with derivative $0$) but not continuously so. You might think about whether "differentiable" or "$C^1$" is the condition you really want.

If you really want just differentiable, I'd start by looking up the counterexample of a function which is differentiable but whose Fourier series doesn't converge. My guess is that it won't be too hard to extend that example to the interior of $\Delta$.

There is a discussion going on in the comments about what the right definition is of a function $f$ on $\bar{\Delta}$ being differentiable at a point $z_0$ of the boundary. We should also keep track of the distinction between definitions which are the analogue of "complex differentiable" and definitions which are the analogue of "real differentiable". The definition I wanted to give was:

(1) There is a complex number $a$ such that $f(z) = f(z_0) + a (z-z_0) + o(|z-z_0|)$ for $z$ in $\Delta$.

We could also give the "real differentiable" analogue of this definition:

(1') There are complex numbers $a$ and $\bar{a}$ such that $f(z) = f(z_0) + a (z-z_0) + \bar{a} (\overline{z-z_0}) + o(|z-z_0|)$ for $z$ in $\Delta$.

I don't know if there is a difference between these.

I haven't fully understood the definition the OP wants to give; he would like to require there to be a differentiable extension of $f$ to an open neighborhood of $z_0$ in $\mathbb{C}$. Again, I can imagine a complex analytic or a real differentiable version of this question. Here are some things I could think of:

(2) There is an open neighborhood $U$ of $z_0$ in $\mathbb{C}$ and a complex differentiable function $g:U \to \mathbb{C}$ so that $g|_{\Delta \cap U} = f|_{\Delta \cap U}$.

But complex differentiable function are $C^{\infty}$ and, as I've already pointed out, the answer is yes for $C^1$ functions. So this definition makes the answer to the original question be "yes".

(2') There is an open neighborhood $U$ of $z_0$ in $\mathbb{C}$ and a real differentiable function $g:U \to \mathbb{C}$ so that $g|_{\Delta \cap U} = f|_{\Delta \cap U}$.

My gut is that this is the same as (1') and, in particular, that my example $(z-1)^2 e^{1/(z-1)}$ probably has a differentiable extension to a neighborhood of $z=1$. But I haven't found one.

Set $g(\theta) = f(e^{i \theta})$.
Lemma The Fourier coefficients of $g$ are $(\ldots, 0,0,0,a_0,a_1,a_2,\ldots)$.
Proof: For $r<1$, we have $\frac{1}{2 \pi} \int_{\theta=0}^{2 \pi} f(r e^{i \theta} )e^{- n i \theta} d \theta = a_n$. Since $f$ is differentiable on $\bar{\Delta}$, it is continuous; since $\bar{\Delta}$ is compact, $f$ is uniformly continuous on $\bar{\Delta}$. So it is valid to interchange integration and limit in $$\lim_{r \to 1^{-}} \int_{\theta=0}^{2 \pi} f(r e^{i \theta} e^{- n i \theta} ) d \theta$$ and conclude that $$\int_{\theta=0}^{2 \pi} f(e^{i \theta} e^{- n i \theta} ) d \theta = a_n.$$ $\square$
Now, the missing part. I want to claim that $g$ is not just differentiable. but $C^1$. If so, then a theorem of Dirichlet states that $g$ is the sum of its Fourier series, as desired.
But now that I think about it, $g$ doesn't need to be $C^1$. Unless I have missed something, $f(z) = (z-1)^2 e^{1/(z-1)}$, as a function on $\bar{\Delta}$, is differentiable at $z=1$ (with derivative $0$) but not continuously so. You might think about whether "differentiable" or "$C^1$" is the condition you really want.