Let $G_1$ and $G_2$ be the groups with the following presentations:

$$G_1=\langle a,b \;|\; (ab)^2=a^{-1}ba^{-1}, (a^{-1}ba^{-1})^2=b^{-2}a, (ba^{-1})^2=a^{-2}b^2 \rangle,$$

$$G_2=\langle a,b \;|\; ab=(a^{-1}ba^{-1})^2, (b^{-1}ab^{-1})^2=a^{-2}b, (ba^{-1})^2=a^{-2}b^2 \rangle,$$

Are these groups torsion-free?

Motivation: In both of these groups $1+a+b$ as an element of the group algebra $\mathbb{F}_2[G_i]$ over the field with two elements is a zero divisor. Thus one has a counterexample for the Kaplansky zero divisor conjecture if one of $G_i$s is torsion-free!