If you grant that $c'=0$ when $c$ is a constant, you can argue as follows:
EDIT: Actually $c'=0$ follows from $1'=0$ using the chain rule, since $c=c\circ 1$.
Let $I(x)=x$. Since $I=I\circ I$, the chain rule gives $(I')^2=I'$, so $I'$ is the characteristic function of a set $A$ of real numbers. Let $T_c(x)=x+c$. Then $T_c'=I'+c'=I'$, and the chain rule applied to $T_c=I\circ T_c$ then implies $I'=(I'\circ T_c)I'$. That means that the set $A$ is either all or nothing. But you can't have $I'$ identically $0$ because that would imply $f'=(f\circ I)'=0$ for all $f$. So $I'=1$ as expected.
Now consider linear functions. If $L_m(x)=mx$, then since $L_m(x+a)-L_m(x)$ is constant, $L_m'(x+a)-L_m'(x)$ is zero. Thus $L_m'$ is a constant depending on $m$, say $h(m)$. The map $h$ is an additive homomorphism, and by the chain rule it is also multiplicative. We also know $h(1)=1$. This implies that $h(m)=m$ for all $m$. (A ring map from reals to reals preserves squares, therefore preserves ordering, therefore is continuous ...)
So $f'$ is what it should be when $f$ is polynomial of degree at most one. Now let $S(x)=x^2$. From $S(x+t)=S(x)+2tx+t^2$ we get $S'(x+t)=S'(x)+2t$, therefore $S'(x)=S'(0)+2x$. But $S'(0)=0$ using $S(-x)=S(x)$. So the derivative of squaring is what it should be.
Now the special case of Leibniz that says $(f^2)'=2ff'$ follows by the chain rule. The general case follows by expressing $fg$ in terms of $f^2$, $g^2$, and $(f+g)^2$.
EDIT: This was all about global functions. But it can be extended. Let me spell out what I hope your axioms are: $f'$ is defined when $f$ is a $C^\infty$ real function whose domain is an open set $U\subset \mathbb R$, and $f'$ is another such function with the same domain. The axioms are
(U) $1_U'=0_U$ where $1_U$ is the constant function on $U$.
(A) $(f+g)'=f'+g'$ where $f$ and $g$ (and $f+g$) have the same domain.
(C) $(f\circ g)'$ is the product of (the restriction to $V$ of) $f'\circ g$ and $g'$, g)'=(f'\circ g)g'$, when $f$ has domain $U$ and $g$ has domain $V$ and $g(V)\subset U$, so that $f\circ g$ has and $f'\circ g$ have domain $V$.
(Z) For every nonempty $U$ there is some $f$ with domain $U$ such that $f'$ is not identically zero.
The arguments that I gave above can be adapted to show then that:
$c_U'=0$ for any constant function on any $U$.
$I_U'=1$ where $I_U$ with domain $U$ is defined by $I_U(x)=x$. (Here you have to mess around with compositions $I_U\circ (I_V+c_V)$.)
So in the end you get the desired localization property, too: the operator commutes with restriction from $U$ to $V\subset U$ by the chain rule, because restriction is composition with $I_V$.