I understand your question that if $X$ is an injective $R\otimes S$ module, then it is an injective $R$ module and the same for projective. In this formulation it seems true, at least if we assume the algebras to have units. Let's start with $X$ being projective. Then there exists an $R\otimes S$ module $Y$ such that $X\oplus Y$ is free.
As $R\otimes S$ is a free $R$-module (the tensor product being over a field), $X\oplus Y$ is also free as an $R$-module, so $X$ is projective as an $R$ module.
Next assume $X$ is $R\otimes S$-injective. Let
0\to M\to N
$$ be an exact sequence of $R$-modules. Then $0\to M\otimes S\to N\otimes S$ is still exact. Now any $R$-module homomorphism $\alpha: M\to X$ induces an $R\otimes S$ module homomorphism $M\otimes S\to X$, which extends to a homomorphism from $N\otimes S$ to $X$. The ensuing map $N\to N\otimes 1\to X$ will be an extension of $\alpha$. So $X$ is injective as $R$ module.