Can a smooth function on the reals form a non-commutative semigroup?

Question

Let $f\colon \mathbb{R}^2 \to\mathbb{R} $ be a smooth function.

Can there exist an algebraic structure $(\mathbb{R}, \cdot)$ such that for $x,y \in \mathbb{R}$, $x \cdot y = f(x,y)$ that is a non-commutative semigroup that is strictly not a monoid or a group?

I can't think of an example, but it seems so unlikely that you can't have such an object.

If not, how does one prove so?

Answer 1

Sure you can, just take $f(x,y)=y$