Is this module over this group algebra projective?

Assume that $G$ is a finite group. Let $k$ be a field. Let $\varepsilon$ be the augmentation $kG\rightarrow k$. Consider the following map

$\varepsilon\otimes id:k[G]\otimes_k k[G]\rightarrow k[G]$

since $k[G]\otimes_kk[G]$ is isomorphic to $k[G\times G]$ this allows us to view $k[G]$ as a $k[G\times G]$-module. My question is: is this module a projective $k[G\times G]$-module?

It is clear when $char\;k\not\mid|G|$ since in this case $k[G\times G]$ is semisimple so everything is projective, but what happens in general? Is that module still projective?

The answer is no. Take for example $A=\mathbb{F}_2[C_2\times C_2]$, where $C_2=\langle c\mid c^2=1\rangle$ is the cyclic group of order 2. I claim that $A$ is indecomposable. Indeed, it is easy to see that the left regular module contains precisely one 1-dimensional submodule $W$ spanned by $$w=(1,1)+(1,c)+(c,1)+(c,c).$$ (Indeed, suppose $x=\alpha(1,1)+\beta(1,c)+\delta(c,1)+\gamma(c,c)$ spans a 1-dimensional submodule. Acting on $x$ with $(1,c)$ shows that $\alpha=\beta$ and $\delta=\gamma$, and acting with $(c,1)$ shows $\alpha=\delta$ and $\beta=\gamma$. Hence, $x=w$.)
On the other hand, take $u=(1,1)+(1,c)$ and $v=(1,1)+(c,1)$. Then $V=\mathrm{span}\{u,v,w\}$ is a 3-dimensional invariant subspace of $_AA$ containing $W$. If it had a compliment, it would be a 1-dimensional submodule different from $W$, which is impossible.
Now, $\mathbb{F}_2[C_2]$ is 2-dimensional, so it cannot be a direct summand of a free $A$-module, as $\dim A=4$ and $A$ is indecomposable.