Here is a partial answer.
Claim. If $\mathcal V$ is an idempotent variety, then any subvariety of $\mathcal V$ that is finitely axiomatizable relative to $\mathcal V$ is axiomatizable by a single identity and the identities of $\mathcal V$.
Apply the claim to the variety of bands. This doesn't explain why all varieties of bands are finitely axiomatizable, but it does explain why only one identity is needed in the finitely axiomatizable cases.
Here is the idea behind the claim.
Suppose that $s=s'$ and $t=t'$ are two identities, where $s, s'$ are terms in, say, two variables, and $t, t'$ are terms in three variables. If $\mathcal V$ is idempotent, then (modulo the identities of $\mathcal V$) the set $\{s=s', t=t'\}$ is equivalent to the single identity
$$
s(t(u,v,w),t(u',v',w'))=s'(t'(u,v,w),t'(u',v',w')).
$$
To explain this, let $M$ be the $2\times 3$ matrix $$ M=\left[ \begin{matrix} u&v&w\\ u'&v'&w' \end{matrix} \right] $$ where the entries are distinct variables. Define a 6-ary term $s\diamond t(M)$ by applying $t$ to the rows of $M$ and then $s$ to the row results. I must explain why $\{s=s', t=t'\}$ is equivalent to $\{s\diamond t(M)=s'\diamond t'(M)\}$.
To derive the diamond identity from the original two, argue from $\{s=s', t=t'\}$ that $$ s\diamond t(M) = s\diamond t'(M) = s'\diamond t'(M). $$
To derive $s=s'$ from the diamond identity, just set variables in $M$ equal along rows, say equal to the first entry: from $s\diamond t(M)=s'\diamond t'(M)$ you obtain $s(u,u')=s'(u,u')$. To derive $t=t'$ from the diamind identity, set variables in $M$ equal along columns. (This paragraph is the part that needs idempotence.)
[Note: lattices are defined with two band operations, and some varieties of lattices are not finitely axiomatizable, so there is still some kind of magic involved in the full result about bands.]