I am looking at the space: $$A:=\left\{f(x)=\sum_{k\in\mathbb{Z}}{a_ne^{inx}}:(a_n)_{n\in\mathbb{Z}}\in l^1(\mathbb{Z})\right\}$$ I want to say the following: if $f\equiv0$, then $a_n=0$ for all $n\in\mathbb{Z}$.
In my opinion, we can solve this problem with the information that $(e^{inx})_{n\in\mathbb{Z}}$ is a basis for $L^2[-\pi,\pi]$, but may you conclude the above result directly form this fact? If not, how do you go further?
You don't need to know that $e^{inx}$ form a basis: the completeness part of the basis definition is irrelevant here. The orthogonality of the exponential functions quickly leads to conclusion. Indeed, fix $m\in\mathbb Z$. For every $k\ge |m|$ we have
$$\int_0^{2\pi} e^{-imx}\sum_{|n|\le k} a_n e^{inx} \,dx = \sum_{|n|\le k}a_n \int_0^{2\pi} e^{i(n-m)x} \,dx = 2\pi a_m \tag{1}$$ (I truncated the sum by $|n|\le k$ to make it finite, so that there is no doubt about the legality of exchanging the order if integration and summation. Just trying to use a minimum theory here.)
The sequence $g_k(x) = e^{-imx}\sum_{|n|\le k} a_n e^{inx}$ converges to $e^{-imx}f(x)=0$ in $L^1(0,2\pi)$ because $$\int_0^{2\pi} |f-g_k| \le 2\pi \sum_{|n|>k}|a_n| \to 0$$ as $k\to\infty$. Thus, the left side of (1) tends to $0$ as $k\to\infty$. But since it's equal to a fixed number $a_m$, it follows that $a_m=0$.