Minimal definition of the derivative

The definition of the Fréchet derivative according to Wikipedia is:

Let $V$ and $W$ be Banach spaces, and $U\subset V$ be an open subset of $V$. A function $f : U \to W$ is called Fréchet differentiable at $x \in U$ if there exists a bounded linear operator $(Df)_x : V\to W$ such that

$$\lim_{h \to 0} \frac{ \| f(x + h) - f(x) - (Df)_x(h) \|_{W} }{ \|h\|_{V} } = 0.$$

I am wondering if perhaps there is some more "axiomatic" interpretation of the derivative. For instance, could we define the derivative as the unique operator on differentiable functions such that

1. The derivative operator is linear. That is, $D(cf) \equiv cD(f)$ and $D(f+g) \equiv D(f)+D(g)$.

2. The chain rule holds. That is, $D(f \circ g)_x \equiv D(f)_{g(x)} \circ D(g)_x$.

3. The product rule holds. That is, if $B : X \times Y \to Z$ is a continuous bilinear operator, then $(DB)_{(x,y)} (u,v) = B(u,y) + B(x,v)$.

4. The derivative operator is the identity on bounded linear operators. That is, if $f: U \to V$ is defined by $f(x) = A(x)$ where $A$ is a bounded linear operator, then for all $x \in U$, $(Df)_x \equiv A$.

Are these rules enough to precisely specify the derivative? Are there some additional rules of this sort that would provide an equivalent definition of the derivative? Are any of these rules redundant?

This does appear to be a valid characterisation of a derivative. Let me try and prove that if $f$ is smooth (say, $C^\infty$, to avoid unnecessary complications), then $D(f)$ is really the derivative of $f$, at least if we additionally assume that $D(c) \equiv 0$ for $c$ - constant map.
First, I claim that we have the product rule, at least in the following form: if $f: V \to \mathbb{R}$ and $g : V \to W$ then $D(fg) = D(f) g + f D(g)$. For this, let $A : V \to V \times V$ be the diagonal map $x \mapsto (x,x)$, let $F(x,x) = (f(x), g(x))$ and finally let $B(t,w) = tw$ for $(t,w) \in \mathbb{R} \times W$. Then, $fg(x)$ can be written as $B \circ F \circ A$. It follows that: $$D(fg)_x = (DB)_{(f(x),g(x))} \circ (DF)_{(x,x)}\circ (DA)_x$$ Looking at compositions of $F$ with projections, it follows that $(DF)_{(x,y)}(u,v) = ((Df)_x(u),(Dg)_y(v))$ (or so it seems to me). From previous axioms we have $(DA)_x = A$, and $(DB)_{(f(x),g(x))} (u,v) = f(x) v + u g(x)$. Combining this all together we conclude that: $$D(fg)_x(v) = f(x) (Dg)_x(v) + (Df)_x(v) g(x)$$ Hence, the usual multiplication rule holds.
Secondly, I claim that if $g: V \to W$ is a map such that $(D_0g)_x = 0$ and $g(0) = 0$, where $D_0$ is the usual derivative, and $x$ is some point in $V$, then $(Dg)_x = 0$. Indeed, if the derivative of $g$ vanishes at $0$, we can write $g$ as $g(x) = f\tilde{g}(x)$ where: $f:\ V \to \mathbb{R}$ is a map with $f(x) = 0$, and $\tilde{g}: V \to W$ is a map with $\tilde{g}(x) = 0$. (At this point, we are using Taylor expansion, and the assumption that $g$ is $C^2$ is needed). From the product rule, it follows that: $$D(g)_x = D(f)_x \tilde{g}(x) + f(x) D(\tilde{g})_x = 0$$.
Finally, we strike the killing blow. Let $g: V \to W$ be any map, and $x$ be a fixed point. We can write $g$ in the form $g(y) = g(x) + (D_0g)_x(y) + h(y)$, where $(D_0h)_x = 0$ and $h(0) = 0$. Now, the previous considerations show that $D_x(h) = 0$, so by additivity (and known behaviour of $D$ on linear and constant functions) we have $(Dg)_x = (D_0g)_x$. Because $x$ was arbitrary, and so was $g$, it follows that $D$ is just the usual derivative.
As for the other possible axioms, I think you could assume the product rule, and work from there. In particular, you can drop the assumption about bilinear maps then. I don't think the other assumptions can be significantly weakened. Without assumption of linearity, it doesn't look like a natural question (at least to me). I also think that without the assumption on what happens for linear operators, $D$ given by something like $Df = P^{-1} (D_0f) P$ for some linear $P$ would work.