The variety of commutative semigroups satisfying $x^3\approx x^4$ is generated by a single algebra 
Let $\mathbf A$ be the algebra given by the following multiplication table
  \begin{array}{c|ccc}
 \style{font-family:inherit}{\cdot} & 0
& 1 & 2 & 3 \\\hline
 0                                       & 0    & 0   & 0 & 0 \\
 1                                       & 0  & 0 & 0 & 1 \\
 2                                       & 0    & 0 & 1 & 2 \\
 3                                       & 0 & 1 & 2 & 3
\end{array}
I need to prove that the variety generated by $\mathbf A$ is exactly the variety of commutative semigroups satisfying $x^3\approx x^4$. 

For one direction, it is easy: since it is checkable that $\cdot$ is commutative and associative in $\mathbf A$ and $\forall x\in A,x^3\approx x^4$, so the variety generated by $\mathbf A$ is in the variety of commutative semigroups satisfying $x^3\approx x^4$. 
For the other direction, I am thinking about decomposing any such commutative semigroup into product of smaller ones, so as to show they lie in $\mathsf{HSP}(\mathbf A)=\cal V(\mathbf A)$ by Birkhoff's Theorem. 
For $1$-element commutative semigroups, there is only one case
\begin{array}{c|ccc}
 \style{font-family:inherit}{\cdot} & 0
 \\\hline
 0                                       & 0
\end{array}
 and it is clearly a subalgebra of $\mathbf A$.
For $2$-element ones, I found that up to isomorphism there should be two types satisfying the identity $x^3\approx x^4$: 
\begin{array}{c|ccc}
 \style{font-family:inherit}{\cdot} & 0
& 1\\\hline
 0                                       & 0    & 0  \\
 1                                       & 0  & 0  
\end{array}
and 
\begin{array}{c|ccc}
 \style{font-family:inherit}{\cdot} & 0
& 1\\\hline
 0                                       & 0    & 0  \\
 1                                       & 0  & 1  
\end{array}
They are also isomorphic to some subalgebras of $\mathbf A$. 
However, starting from $3$-element commutative semigroups, it seems hard to characterise types satisfying the identity. On the other hand, the subalgebras in $\mathbf A$ of $3$ elements seem to contain only $\{0,1,2\}$ and $\{0,1,3\}$. 
So I wonder whether there is a better way to prove this direction? 
Thank you very much for any help! 

Let $V$ be the variety of all commutative semigroups
satisfying $x^3\approx x^4$. Let
$V({\mathbf A})$ be the subvariety generated by ${\mathbf A}$.
If $V({\mathbf A})$ is a proper subvariety of $V$, then
there is a law that holds in ${\mathbf A}$ that does
not hold throughout $V$. Using the identities of $V$, we may
reduce any such law to one of the form
$s\approx t$ where
$s = x_1^{a_1}\cdots x_k^{a_k}$,
$t = x_1^{b_1}\cdots x_k^{b_k}$,
and $a_i, b_i\in\{0,1,2,3\}$ for all $i$.
Here a power of the form $x_i^0$, with exponent $0$, 
should be interpreted as the identity
element of $\mathbf A$, which is $3$.
Since $s\approx t$ does not hold in $V$, there must be
some index where the variables in
these words have different exponents,
say $a_j\neq b_j$.
Substitute the identity element $3\in A$
for all variables except the $j$th,
and substitute $2$ for $x_j$. You obtain from $s\approx t$ that
$2^{a_j}=2^{b_j}$.
But the possible powers of $2$ are all distinct:
$2^0 = 3, 2^1 = 2, 2^2 = 1, 2^3 = 0$. This makes 
it impossible to have
$a_i, b_i\in\{0,1,2,3\}$, $a_j\neq b_j$, and
$2^{a_j}=2^{b_j}$. 
