How to prove when $R$ is not a commutative ring, $M\otimes_R N$ is not a module

Question

I know how to prove when $R$ is a commutative ring, $M\otimes_R N$ is a module, but how to prove when $R$ is non-commutative ring, we cannot give any module structure on $M\otimes_R N$? I see many answers say that we cannot define $r(m\otimes n)$= $m\otimes rn$ because it will broke the bilinearity. However, it is not the only way to define $$h: R\times (M\otimes_R N)\to (M\otimes_R N),$$ right? If I want to show $M\otimes_R N$ is not a module, I need to prove any function $h$ cannot satisfy axioms of module. I think it can be proved by contradiction: assume that $h$ is a function makes $M\otimes_R N$ become a module, then for each $r\in R$, $h_r$ is a homomorphism from $M\otimes_R N$ to $M\otimes_R N$, and correspond a bilinear function from $M\times N$ to $M\otimes_R N$. But I don't know how to do next, can you give me some hints to help me finish the proof or give me another proof show that when $R$ is not a commutative ring, $M\otimes_R N$ is not a module? Thanks!