Let V be the variety of algebras (A,\ \cdot) satisfying

Step 1
To save typing, I'll write ${\displaystyle x\cdot y}$ and ${\displaystyle xy}$ and $\approx$ as ${\displaystyle =}$.
Bergman gives in part (a) a list of identities which follow from the defining identities of ${\displaystyle V}$ These identities are very helpful in solving part (b)!
1) ${\displaystyle \left(xy\right)\left(zw\right)=\left(xz\right)\left(yw\right)}$
2) ${\displaystyle x\left(yz\right)=\left(xy\right)\left(xz\right)}$
3) ${\displaystyle \left(yz\right)x=\left(yx\right)\left(zx\right)}$
4) ${\displaystyle y\left(xy\right)=\left(yx\right)y}$
5) ${\displaystyle \left(yx\right)x=xy}$
For completeness, I'll give proofs of these five identities inside the spoiler blocks.
1) ${\displaystyle \left(xy\right)\left(zw\right)=\left(\left(zw\right)y\right)x=\left(\left(yw\right)z\right)x=\left(xz\right)\left(yw\right)}$
2) ${\displaystyle x\left(yz\right)=(\times )\left(yz\right)=\left(xy\right)\left(xz\right)}$
3) ${\displaystyle \left(yz\right)x=\left(yz\right)(\times )=\left(yx\right)\left(zx\right)}$
4) ${\displaystyle y\left(xy\right)=\left(yx\right)\left(yy\right)=\left(yx\right)y}$
5) ${\displaystyle \left(yx\right)x=(\times )y=xy}$
Ok, now we add the additional defining identity of ${\displaystyle W:6.y\left(xy\right)=x}$. Note that in conjunction with identity 4 above, we also have 7. ${\displaystyle \left(yx\right)y=x}$
Our task is to understand ${\displaystyle F_W(a,b)}$, the free algebra in ${\displaystyle W}$ on two generators a and b. I'm using a and b for the generators so as not to be confused with the variables x and y used in the identities above. Let's try to find all its elements.
Recall that every element of the free algebra ${\displaystyle F_W(a,b)}$ is an equivalence class of terms in the generators a and b. The terms can be built up in levels, where the terms at Level 0 are the generators and the terms at Level ${\displaystyle (n+1)}$ are the generators and the products of two terms at Level ${\displaystyle (\le n)}$.
Level 0: a, b
Level 1: a, b, aa, ab, ba, and bb.
Note that we can eliminate aa and bb since they are redundant: ${\displaystyle aa=a}$ and ${\displaystyle \mathbf{=}b}$
Now it turns out that any term at Level 2 (any product of a, b, ab, and ba) can be shown to be equivalent to a term at Level 1, using the identities above. They multiply according to the following Cayley table:
$$\begin{array}{ccccc}& a& b& ab& ba\\ a& a& ab& ba& b\\ b& ba& b& a& ab\\ ab& b& ba& ab& a\\ ba& ab& a& b& ba\end{array}$$
I've hidden the derivations in the spoiler blocks below.
${\displaystyle a\left(ab\right)=a\left(\left(aa\right)b\right)=a\left(\left(ba\right)a\right)=ba}$ by 6. Similarly, ${\displaystyle b\left(ba\right)=ab}$.
${\displaystyle b\left(ab\right)=a}$ by 6. Similarly, ${\displaystyle a\left(ba\right)=b}$.
${\displaystyle \left(ab\right)a=b}$ by 7. Similarly,