There's nothing particularly special about $kG$ as far as this property goes: in *any* finite ring $R$, every element is either a unit or a zero divisor, and is a unit if and only if some power is the identity. (Proof: Given an element $x\in R$, if the left-multiplication map $L_x : R \to R$ is not injective, then there are two elements $r_1\neq r_2$ with $xr_1=xr_2$, which implies that $x(r_1-r_2)=0$, and $x$ is a zero divisor. Likewise, if the right-multiplication map $R_x$ is not injective, then $x$ is a zero-divisor. So assume both $L_x$ and $R_x$ are injective. Since $R$ is finite, it follows that $L_x$ and $R_x$ are also surjective, so that there exist $r_1,r_2\in R$ with $r_1x=1=xr_2$. Then $r_1=r_11=r_1xr_2=1r_2=r_2$, and $r_1=r_2$ is an inverse to $x$, so $x$ is a unit.)

As far as characterizing the units and zero divisors of $kG$, we can consider the Wedderburn decomposition of $kG$ as a direct product of full matrix rings over division rings:
$$kG \cong M_{n_1}(D_1) \times \cdots \times M_{n_s}(D_s)$$
Here, the divison rings $D_i$ are finite, hence they are fields (by Wedderburn's little theorem) and must be extensions of the base field $k$ and hence have size $|D_i|=q^{m_i}$ for some $m_i\geq 1$, where $|k|=q$. So we may rewrite the Wedderburn decomposition,
$$kG \cong M_{n_1}(\mathbb F_{q^{m_1}}) \times \cdots \times M_{n_s}(\mathbb F_{q^{m_s}})$$

The units can then be characterized as the group
$$(kG)^* \cong GL_{n_1}(\mathbb F_{q^{m_1}}) \times \cdots \times GL_{n_s}(\mathbb F_{q^{m_s}})$$