This is true, $\frac{\partial}{\partial S}C_\text A(S,0)\in [0,1]$ and that the difference is bound above.
Proof:
Suppose $S\ge 0$ follows a local volatility process
$$dS(t) = \sigma(S(t),t)S(t)\,dB(t)$$
where $B(t)$ is the standard Brownian motion. We can extend this dynamics later to account for a stochastic volatility process.
Because $C_\text E(S,t_1^-)=C_\text E(qS,t_1^+)$, $\frac{\partial}{\partial S}C_\text E(S,t_1^-)=q\frac{\partial}{\partial x}C_\text E(x=qS,t_1^+)<q<1,\,\forall t<T$,
$S-K=C_{\text E}(S,t_1^-)$ has a unique solution for $S$. We denote it by $S_c$. Because the value of a call, either American or European, such as $C_\text E(S,t)$, is convext with respect to stock $S$, $\min_{S\ge S_c}\frac{\partial}{\partial S}C_\text E(S,t_1^-)=\frac{\partial}{\partial S}C_\text E(S=S_c,t_1^-)$.
Let $\Theta$ be the Heaviside step function. At time $t<s$, both calls, and their difference $C=C_{\text A}-C_{\text E}$ due to the linearity thereof, satisfy
$$\bigg[\frac{\partial}{\partial t}+\frac12(\sigma S)^2\frac{\partial^2}{\partial S^2}\bigg]C(S,t)=0,$$
$$C(S,t=t_1^-)=\big(S-K-C_\text E(S,t_1)\big)\Theta(S-S_c).$$
Take partial derivative of the above partial differential equation over $S$ and $C_1:=\frac{\partial}{\partial S}C$. We have
$$\bigg[\frac{\partial}{\partial t}+uS\frac{\partial}{\partial S} +\frac12(\sigma S)^2\frac{\partial^2}{\partial S^2}\bigg]C_1(S,t)=0, \tag1\label{1}$$
$$C_1(S,t=t_1^-)=\Big(1-\frac{\partial}{\partial S}C_\text E(S,t_1)\Big)\Theta(S-S_c),$$
where $u:=\frac1{2S}\frac{\partial}{\partial S}(\sigma S)^2$.
It can be shown by repeating the same methodology here that
$$0\le C_1(S,t_1^-)\le1-\frac{\partial}{\partial S}C_\text E(S=S_c,t_1^-).$$
By the strong maximum principle of elliptical and parabolic partial differential equation (c.f. also the Wikipedia article),
$$0=\inf_S C_1(S,t_1^-)\le C_1(S,t)\le\sup_S(S,t_1^-)=1-\frac{\partial}{\partial S}C_\text E(S=S_c,t_1^-), \quad\forall t<t_1. \tag2\label2$$
An alternative to the last step is the Feynman-Kac formula.
The solution to Equation $\eqref1$ is
$$C_1(S,t)=\mathbf E\big[C_1(S(t_1^-),t_1^-)\big|S(t_1^-)=S\big]$$
under the probability measure such that $S(t)$ is a new Ito process
$$dS=uSdt+\sigma S\,dB.$$
Equation $\eqref2$ is then obvious.
To recapitulate, we have proved -- more than what is sought by the question -- that
$$0\le\frac{\partial}{\partial S}\big(C_{\text{A}}(S,t)-C_{\text{E}}(S,t)\big)\le1-\frac{\partial}{\partial S}C_\text E(S=S_c,t_1^-), \quad\forall t<t_1,$$
which is the desired result for the difference.
Apply the same method, we can show
\begin{align*}
& C_\text A(S,T) = C_\text E(S,T) = (S-K)_+ \\
\implies & \frac{\partial}{\partial S}C_\text A(S,T) = \frac{\partial}{\partial S}C_\text E(S,T) = \Theta(S-K)\in [0,1] \\
\implies & \frac{\partial}{\partial S}C_\text A(S,t_1^+) = \frac{\partial}{\partial S}C_\text E(S,t_1^+) \in [0,1]
\end{align*}
and thus
$$\frac{\partial}{\partial S}C_\text A(S,t)\in [0,1], \quad\forall t<t_1.$$
$$\tag*{$\square$}$$
