If $G$ is a $p$-group, this is almost never true (and I believe but may be wrong that for general $G$, it's completely governed by the $p$-Sylow). In that case, we can understand modules in terms of the support variety $V={\rm Proj} H^{2*}(G,k)$, the projective variety associated to the (even-dimensional) cohomology ring of $G$. To a finite $kG$-module $M$ we can associate its support in $V$, namely the support of Ext(M,M) as a graded $H^{2*}(G,k)$-module. This support is a closed subset of $V$, and conversely every closed subset is the support of some finite $kG$-module. Finally, the support of $M\otimes N$ is the intersection of the support of $M$ and the support of $N$, and a module is projective iff its support is empty. Thus $M\otimes N$ is projective iff $M$ and $N$ have disjoint support.
Thus unless $V$ is just a single point, it is possible to have non-projective modules whose tensor product is projective. For $V$ to be a point, the cohomology of $G$ must be a polynomial ring in one variable, up to nilpotent elements. By Quillen's theorem, this is the case iff all elementary abelian subgroups of $G$ are conjugate and rank 1.
In particular, it is true for cyclic groups, but otherwise it is almost always false. There's a simple argument to see directly that it holds for cyclic groups of order $p$: in that case, $kG$ can be identified with $k[x]/x^p$, and every indecomposable module is of the form $M_i=k[x]/x^i$ for some $i\leq p$. Such a module $M_i$ is projective iff $i=p$. If $M_i$ and $M_j$ are not projective, then $M_i\otimes M_j$ has dimension $ij$, which is not divisible by $p$. Thus $M_i \otimes M_j$ cannot be a sum of copies of $M_p$ and is hence not projective.