The answer to the first question is no. Let $G$ be the group $C_2 \times C_2$ and let $k$ be a field of characteristic 2. Then $kG \cong k[x,y]/(x^2, y^2)$. Let $M = k[x]/(x^2)$ with $y$ acting trivially, and let $N = k[y]/(y^2)$ with $x$ acting trivially. Then neither $M$ nor $N$ is projective (for modules over a $p$-group in characteristic $p$, projective is the same as free), but $M \otimes_k N \cong kG$.

Given this example, it's also hard to imagine what a good answer for the second question could be.

The answer to the first question is no. Let $G$ be the group $C_2 \times C_2$ and let $k$ be a field of characteristic 2. Then $kG \cong k[x,y]/(x^2, y^2)$. Let $M = k[x]/(x^2)$ with $y$ acting trivially, and let $N = k[y]/(y^2)$ with $x$ acting trivially. Then neither $M$ nor $N$ is projective (for modules over a $p$-group in characteristic $p$, projective is the same as free), but $M \otimes_k N \cong kG$.

Edit: I think that if the Sylow $p$-subgroup of $G$ is cyclic, then you can use Avrunin and Scott's tensor product theorem to prove that whenever $M \otimes_k N$ is projective, then so is either $M$ or $N$, but that seems like a big tool to use for this.