Good questions! To bump the discussion of torsion out of the comments: the group of units of $\mathbb{F}_2[P]$ is torsion-free. Suppose we have $q$-torsion and factor $0 = x^q - 1 = (x - 1)(x^{q-1} + \dots + x + 1)$. For $q = 2$ this immediately contradicts the zero divisor conjecture unless $x = 1$. For odd $q$ we use the fact that $x$ must map to $1 \in \mathbb{F}_2$ under the augmentation map to see that $1 + x + \dots + x^{q-1} \neq 0$ and get the contradiction.

Any non-trivial unit in $\mathbb{Z} G$ for $G$ torsion-free (if such a thing exists!) is infinite order by a theorem of Sehgal.

Sehgal, Sudarshan K., Certain algebraic elements in group rings, Arch. Math. 26, 139-143 (1975). ZBL0322.20002.

Good questions! To bump the discussion of torsion out of the comments: the group of units of $\mathbb{F}_2[P]$ is torsion-free. Suppose we have $q$-torsion and factor $0 = x^q - 1 = (x - 1)(x^{q-1} + \dots + x + 1)$. For $q = 2$ this immediately contradicts the zero divisor conjecture unless $x = 1$. For odd $q$ we use the fact that $x$ must map to $1 \in \mathbb{F}_2$ under the augmentation map to see that $1 + x + \dots + x^{q-1} \neq 0$ and get the contradiction.

Any non-trivial unit in $\mathbb{Z} G$ for $G$ torsion-free (if such a thing exists!) is infinite order by a theorem of Sehgal.

Sehgal, Sudarshan K., Certain algebraic elements in group rings, Arch. Math. 26, 139-143 (1975). ZBL0322.20002.

Overdue update: my paper now has a corollary showing that the group of units has free subgroups and is not finitely generated. Furthermore, Murray has given non-trivial units in all positive characteristics (see arXiv).