Why is every variety of bands determined by a single identity? A band is a semigroup where every element is idempotent: $a a = a$.  A collection of bands is a variety  if it is closed under taking subobjects (subsemigroups), quotient objects (images of homomorphisms) and products (including infinite products).  As usual in universal algebra, any variety of bands is defined by a collection of identities.  For example, there's a variety of semilattices, which are the bands obeying the identity
$$ a b = b a $$
for all $a,b$.  
You can define a variety of bands using more than one identity.  But amazingly --- to me, at least --- every variety of bands can in fact be defined using just one identity! 
(That is, one identity in addition to associativity $(a b) c = a (b c)$ and the idempotence law $a a = a$.)
This was shown here:


*

*Charles Fennemore, All varieties of bands, Semigroup Forum 1 (1970), 172-179.


He even showed that there are exactly $8 + 10(n-2)$ varieties of bands that are determined by an identity involving $n$ variables, for $n \ge 2$... and he has a method for listing these identities explicitly.
The set of varieties of bands is partially ordered, based on whether one identity implies another.  It's actually a lattice, as usual in universal algebra, and it seems that Fennemore described the lattice operations explicitly in terms of his chosen identities.
A small portion of this lattice is shown on Wikipedia:

I'm wondering if there's a good explanation of 'why' any variety of bands can be defined using just one identity.  Fennemore's proof is not easy for me to follow.  Are there are other varieties, besides bands, that have this property?
 A: There is likely a good technical explanation to be offered; I'm not feeling up to
producing one.  I do have some handwaving intuition that might help.
Your example is based in a finite language/signature (one binary operation symbol),
so I will restrict myself to finite functional languages, even though some
things might hold for countable languages.  The subvariety lattice for a
given finite signature corresponds in a dual fashion to the lattice of
closed (first-order) equational theories in that signature, which in turn
correspond to the lattice of certain (fully-invariant, I think) congruences
in the free term algebra on countably many generators.  This lattice can be
pretty woolly, but you've found a nice sublattice of it which corresponds to
the subvarieties of bands.  Further, this nice sublattice has a property of
representation that you pointed out: relative to the congruence that
defines or describes all bands, any further congruence containing it is
principally and singularly defined, needing just one pair of terms to be
defined to generate the rest of it.
Why is this?  In the case of bands, there is a simple enough set up: reduce all
terms using associativity to a nice normal form (semigroup words), then use
idempotency to further reduce possibilities.  If I recall correctly, idempotency
really cuts down on the available distinct words: in particular, bands are a locally
finite variety, so for each n there will be only finitely many distinct terms in
n variables to choose from.
Why do you get to use exactly one identity?  This is less clear to me, but recall
that an identity in n variables implies (by substitution) several identities in
m variables for m less than n.  There may be a way in the case of bands to order
the relatively free terms by implicational strength, so that the result is a linear
(or not very wide partial)
order.  Then any collection of identites can be reduced by some process to one.
The last part involved a bit of arm waving as well.  I want to emphasize the point
that if you set up an initial base of identities in your theory, you can maneuver
into a position where there aren't many more steps to x=y, or triviality.  If you
start high enough (in the lattice of fully invariant congruences of the term algebra), I think you can carve out many theories which represent
their completions in such a nice manner.
A: 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 diamond 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.]
