Torsion-freeness of two groups with 2 generators and 3 relators and Kaplansky Zero Divisor Conjecture

Denote $x=ab$, $y=a^{-1}ba^{-1}$. Then the first two relations of the first group are $x^2=y$, $y^2=(yx)^{-1}$. This implies $x^4=x^{-3}$ or $x^7=1$. So the group has torsion. I leave the second group as an exercise for the others.

Similar to the answer of Mark Sapir, for the second group let $x=ba$ and $y=b^{-1}ab^{-1}$. From the second relation we have $y^2=x^{-1}y^{-1}$. So, from the first relation we have $x=y^4$. These relations implies that $y^4=y^{-3}$ or $y^7=1$.