Does Function to Function Multiplication and Convolution, Follow Laws of Convolution
In a nutshell, you are asking whether the following statement is true:

$$[x(n)\cdot y(n)] *h(n) = [x(n)*h(n)] \cdot [y(n)*h(n)]$$

In words, if the convolution operator is distributive with multiplication. 

The short answer is no.

An easy example would be to think of two sequences $h(n)$ and $x(n)$ such that their convolution is $0 \ \forall n$ (if you can't think of any, check this out). We can see that in this case the right side of the equality given above would be $0$ no matter what $y(n)$ is, but the left one would not (at least not for all $y(n)$).