The second theorem below was first proved by Wiener (1933). Later Gelfand (1941) found a wonderful algebraic proof based on Banach algebras, which I included since it is so simple and elegant.

Theorem (Gelfand-Mazur) If $A$ is a complex Banach algebra with unity $e$ whose nonzero elements are invertible, then $A\cong\mathbb{C}$.

Proof (sketch). For any $a\in A$ there exists $\lambda(a)\in\mathbb{C}$ such that $a=\lambda(a)e$: otherwise $a-\lambda e$ would be invertible for any $\lambda\in\mathbb{C}$ and, choosing any $\phi\in A^*$ (the dual space) such that $\phi(a^{-1})\neq 0$, the holomorphic map $\lambda\mapsto\phi((a-\lambda e)^{-1})$ would contradict Liouville's theorem. Clearly $a\mapsto\lambda(a)$ defines an isomorphism with $\mathbb{C}$.

Theorem (Wiener). If $f:S^1\to\mathbb{C}\setminus\{0\}$ has the form $f(e^{i\theta})=\sum_{n\in\mathbb{Z}}c_n e^{in\theta}$ with $\sum|c_n|<\infty$, then $\frac{1}{f}$ has the same form (i.e. $\frac{1}{f(e^{i\theta})}=\sum_{n\in\mathbb{Z}}c_n'e^{in\theta}$ with $\sum|c_n'|<\infty$).

Proof (Gelfand). The set of functions $g(e^{i\theta})=\sum_{n\in\mathbb{Z}}a_n e^{in\theta}$ with $\sum|a_n|<\infty$ forms a commutative Banach algebra $B$ with the norm $\|g\|:=\sum|a_n|$ (and multiplicative identity $1$). The thesis amounts to show that $f$ is invertible in $B$.

If this does not happen, then $f$ is contained in some maximal ideal $M$, which has to be closed by maximality (since invertible elements form an open set). By Gelfand-Mazur theorem $B/M\cong\mathbb{C}$, so there exists a homomorphism $\phi:B\to\mathbb{C}$ satisfying $\phi(f)=0$. Notice that, for any $b\in B$, $|\phi(b)|\le\|b\|$: indeed, whenever $|\lambda|>\|b\|$ the element $b-\lambda\cdot 1=-\lambda(1-\lambda^{-1}b)$ is invertible in $B$ and thus $\phi(b)-\lambda=\phi(b-\lambda\cdot 1)\neq 0$. In particular, $\phi$ is continuous.

Let $h(e^{i\theta}):=e^{i\theta}$. Observe that $h\in B$ is invertible and that, for any $n\in\mathbb{Z}$, $\|h^n\|=1$. So $|\phi(h)|^n=|\phi(h^n)|\le 1$ for all $n$, i.e. $\phi(h)\in S^1$. Thus we arrive at the contradiction $$ 0=\phi(f)=\sum c_n\phi(h^n)=\sum c_n\phi(h)^n=f(\phi(h)). $$