I suspect that it's not possible to approximate continuous functions by singular functions (continuous functions with 0 derivative a.e.) in $A(\mathbb{T})$. I definitely can't prove this, but I'll give some reasons below why it seems unlikely.
Heuristically, I would expect that the lack of smoothness of (non-constant) singular functions would lead to only weak decay of the Fourier coefficients. I might even hazard a guess that most non-constant singular functions aren't in $A(\mathbb{T})$ at all.
To go a little further in this direction, let's let $f$ be a non-constant singular function on $[0, 2 \pi]$, and let $\mu$ be the measure that integrates to $f$, i.e. $f(x) = \int_{0}^{x} d\mu(y)$. Then
$$
\hat{f}(\xi)
= \int_{0}^{2 \pi} e^{-i \xi x} f(x) \, dx
= \frac{e^{-i \xi} (f(0) - f(2 \pi))}{i \xi}
+ \frac{\hat{\mu}(\xi)}{i \xi}.
$$
The first term on the right is really what we're up against: if it dominates, the Fourier coefficients of $f$ don't decay fast enough to be summable. This means we can't even afford that $\hat{\mu}(\xi) \to 0$ as $|\xi| \to \infty$. (Note that if $f(0) = f(2 \pi)$ then the first term vanishes, and we may be okay for $A(\mathbb{T})$. But such functions won't, I think, help us to construct approximations of arbitrary continuous functions, since generally we would need to piece together up-and-down movements.)
But at this point, making widely applicable statements about $\hat{\mu}$ gets fuzzier, from what I can tell. If $\mu$ is a Cantor measure, it appears that $\hat{\mu}$ typically has some positive decay rate, although for certain dissection ratios there may not be decay. (See Fourier decay rate of Cantor measures.) Perhaps the non-decay situation would an avenue to explore if you think the approximation in $A(\mathbb{T})$ ought to be possible.
Beyond that, we're wading into the territory of measures that are continuous and singular with respect to Lebesgue measure, but in some sense not fractal-y. (Well, I suppose there are variants of fractal measures beyond the basic Cantor measures discussed above; that would be an intermediate area to check out.) I'm not even sure how to make that more precise, so I've run out of steam.
