This is to summarize what were discussed in the comments, so the title will not be listed as unanswered.

The linear subspace $S$ of $c_0(\mathbb{Z})$ is equal to the convolution product of two copies of $\ell^2(\mathbb{Z})$.

More precisely, $\lbrace a_n \rbrace$ is in $S$ if and only if there exist two sequences $\lbrace b_n \rbrace$ and $\lbrace c_n \rbrace$ in $\ell^2(\mathbb{Z})$ such that $$a_n=\sum_{k=-\infty}^\infty b_k c_{n-k} $$ for all $n$.

This follows since every function in $L^1[0,2\pi]$ is a product of two functions in $L^2[0,2\pi]$, and that for any functions $f,g$ in $L^2[0,2\pi]$ one has $$\frac{1}{2\pi}\int_{-\pi}^\pi f(x)g(x)e^{-inx}dx=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)\overline{h(-x)}e^{-inx}dx=\sum_{k=-\infty}^\infty \hat{f}(k) \hat{h}(n-k)$$ where $h(x)=\overline{g(-x)}$.

(One also uses that the mapping that maps each $f$ in $L^2[0,2\pi]$ to its Fourier coefficient sequence is an isomorphic isometry.)

This is to summarize what were discussed in the comments, so the title will not be listed as unanswered.

The linear subspace $S$ of $c_0(\mathbb{Z})$ is equal to the convolution product of two copies of $\ell^2(\mathbb{Z})$.

More precisely, $\lbrace a_n \rbrace$ is in $S$ if and only if there exist two sequences $\lbrace b_n \rbrace$ and $\lbrace c_n \rbrace$ in $\ell^2(\mathbb{Z})$ such that $$a_n=\sum_{k=-\infty}^\infty b_k c_{n-k} $$ for all $n$.

This follows since every function in $L^1[0,2\pi]$ is a product of two functions in $L^2[0,2\pi]$, and that for any functions $f,g$ in $L^2[0,2\pi]$ one has, by Parseval identity, $$\frac{1}{2\pi}\int_{-\pi}^\pi f(x)g(x)e^{-inx}dx=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)\overline{h(x)}e^{-inx}dx=\sum_{k=-\infty}^\infty \hat{f}(k) \hat{g}(n-k)$$ where $h(x)=\overline{g(x)}$.

(One also uses that the mapping that maps each $f$ in $L^2[0,2\pi]$ to its Fourier coefficient sequence in $\ell^2(\mathbb{Z})$ is a surjective isomorphic isometry.)

This is to summarize what were discussed in the comments, so the title will not be listed as unanswered.

The linear subspace $S$ of $c_0(\mathbb{Z})$ is equal to the convolution product of two copies of $\ell^2(\mathbb{Z})$.

More precisely, $\lbrace a_n \rbrace$ is in $S$ if and only if there exist two sequences $\lbrace b_n \rbrace$ and $\lbrace c_n \rbrace$ in $\ell^2(\mathbb{Z})$ such that $$a_n=\sum_{k=-\infty}^\infty b_k c_{n-k} $$ for all $n$.

This follows since every function in $L^1[0,2\pi]$ is a product of two functions in $L^2[0,2\pi]$, and that for any functions $f,g$ in $L^2[0,2\pi]$ one has $$\frac{1}{2\pi}\int_{-\pi}^\pi f(x)\overline{g(-x)}e^{-inx}\,dx=\sum_{k=-\infty}^\infty \hat{f}(k) \hat{g}(n-k).$$

(One also uses that the mapping that maps each $f$ in $L^2[0,2\pi]$ to its Fourier coefficient sequence is an isomorphic isometry.)

This is to summarize what were discussed in the comments, so the title will not be listed as unanswered.

The linear subspace $S$ of $c_0(\mathbb{Z})$ is equal to the convolution product of two copies of $\ell^2(\mathbb{Z})$.

More precisely, $\lbrace a_n \rbrace$ is in $S$ if and only if there exist two sequences $\lbrace b_n \rbrace$ and $\lbrace c_n \rbrace$ in $\ell^2(\mathbb{Z})$ such that $$a_n=\sum_{k=-\infty}^\infty b_k c_{n-k} $$ for all $n$.

This follows since every function in $L^1[0,2\pi]$ is a product of two functions in $L^2[0,2\pi]$, and that for any functions $f,g$ in $L^2[0,2\pi]$ one has $$\frac{1}{2\pi}\int_{-\pi}^\pi f(x)g(x)e^{-inx}dx=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)\overline{h(-x)}e^{-inx}dx=\sum_{k=-\infty}^\infty \hat{f}(k) \hat{h}(n-k)$$ where $h(x)=\overline{g(-x)}$.

(One also uses that the mapping that maps each $f$ in $L^2[0,2\pi]$ to its Fourier coefficient sequence is an isomorphic isometry.)