When do Leibniz-like rules lead to unique linear operators

Question

Background

Usually one defines differentiation in terms of limits, and then shows that differentiation satisfies the Leibniz (product) rule, $$\frac{d}{dx}(f \cdot g) = f\frac{dg}{dx} + g \frac{df}{dx},$$ as a consequence.

However, in this mathoverflow post I was surprised to find that one can go the other way around and use the Leibniz rule to define the derivative in a more algebraic/axiomatic manner. In particular, there exists a unique linear operator $L: C^\infty(\mathbb{R}) \rightarrow C^\infty(\mathbb{R})$ satisfying the following three properties:

1. $L(1) = 0$
2. $L(x) = 1$
3. $L(f \cdot g) = f \cdot L(g) + g \cdot L(f),$

and so we can use these properties as the definition of the derivative operator instead.

Question

This naturally leads to the following question: if we replace the Leibniz rule with some different polynomial of the relevant quantities, $$L(f \cdot g) = \text{polynomial}\left(f,g,L(f),L(g)\right),$$ for example, $$L(f \cdot g) = g^3 \cdot L(f) \cdot L(g)^2 + f^2 \cdot g,$$

then under what conditions do the analogous three properties,

1. $L(1) = 0$
2. $L(x) = 1$
3. $L(f \cdot g) = \text{polynomial}\left(f,g,L(f),L(g)\right),$

also lead to a unique linear operator?

If so, are there specific examples of polynomials that lead to familiar linear operators (other than the derivative)?

Works on polynomials?

So far my thinking is as follows: given these three rules, the action of $L$ is uniquely defined on any polynomial function since we can recursively apply the "Leibniz-like" rule (3.) to break things down until we reach either of the base cases (1., 2.). Then using the Weierstrauss theorem on a compact set we should be able to extend the action of $L$ from polynomials to smooth functions by density.

However, it is not clear to me that the operator remains unique on the larger space of smooth functions on a compact set, nor is it clear how to make the argument work on all of $\mathbb{R}$. More importantly, it's not clear what the "meaning" of such an operator would be (in the same manner as Liebniz -> derivative).