Assume that a Banach algebra $B$ has such a basis $H$. Take any $h\in H$. Consider the element $x=ah+a^2h^2+a^3h^3+\dots$ where $a>0$ is chosen small enough to make the series converge in $B$. Assume $x=\sum_{j=1}^n c_jh_j$ for some $h_j\in H$. Write
$$
x=ah+a^2h^2+\dots+a^mh^m+a^mh^mx=ah+a^2h^2+\dots+a^mh^m+\sum_j c_ja^m(h^mh_j)
$$
If all the basis elements $h^k\in H$, $k=1,\dots,m$ are distinct, then the elements $h^mh_j\in H$ in the last sum can cancel only $n$ of them and we get a representation for $x$ of length $\ge m-n>n$ if $m>2n$. Thus some two of the powers must coincide and, thereby, $h^k$, $k\ge 1$ span only a finite dimensional space. Since it is true for every $h\in H$, we conclude that the powers of every element of $B$ span only a finite-dimensional space. This excludes any "decent" infinite-dimensional Banach algebra ($c_0$ in particular).
However something survives. The most degenerate example is the following. Take some Banach algebra, say $B=c_0$, an idempotent element $y\in B$ (of norm $1$, say, though it doesn't matter), say $y=(1,0,0,0,\dots)$, some linear multiplicative functional $\psi$ on $B$ such that $\psi(y)=1$, which would be the evaluation of the first coordinate in that particular example, and redefine the product in $B$ to
$$
ab=\psi(a)\psi(b)y
$$
(this is not a very exciting product, of course, but it is a product nevertheless).
Now just take any maximal linearly independent set containing $y$ in $B\setminus\operatorname{ker}\psi$ and normalize its elements $h$ by replacing them by $\frac 1{\psi(h)}h$. You have your basis.
