= Solution
Use the basis of $D$-conjugation orbit sums in $kG$. A singleton orbit is exactly an element of $C_G(D)$, so these basis vectors span the subring $kC_G(D)$. For a nonsingleton orbit with representative $x$, its stabilizer is the proper <subgroup> $E=C_D(x)<D$, and its orbit sum is $\operatorname{Tr}_E^D(x)$. Thus every nonsingleton basis vector belongs to the sum of proper-subgroup transfer images.
Conversely the coefficient at a $D$-fixed basis element $x\in C_G(D)$ in $\operatorname{Tr}_E^D(a)$ is $[D:E]a_x=0$ in characteristic $p$ whenever $E<D$. Hence the two spans intersect trivially, giving
$$
\boxed{(kG)^D=kC_G(D)\oplus I_D^{\mathrm{prop}},\qquad
I_D^{\mathrm{prop}}=\sum_{E<D}\operatorname{Tr}_E^D((kG)^E).}
$$
For $u\in(kG)^D$, multiplication by a transfer element satisfies $u\operatorname{Tr}_E^D(a)=\operatorname{Tr}_E^D(ua)$ and $\operatorname{Tr}_E^D(a)u=\operatorname{Tr}_E^D(au)$. Therefore the second summand is a two-sided <ideal>, as required; it need not be nilpotent in this general setting.
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