Another thought: (I use $(n,a,b)$ instead of $(C,\gamma,p)$)
Claim 1 $\hspace{5ex}$ Let $a_n,b_n>0$ with $a_n,b_n=\mathcal O(n^p)$ for some $p\in\mathbb R$ and $a,b>0$. Then $\lim_{n\to+\infty}-\frac{1}{n}\ln\left(1+(a_ne^{-an}-1)(1-b_ne^{-bn})\right)=\min\{a,b\}.$
Indeed $\hspace{5ex}$ For $a\leq b$, (the other case is similar) $\begin{aligned}-\frac{1}{n}\ln\left(1+(a_ne^{-an}-1)(1-b_ne^{-bn})\right)&=a-\frac{\ln a_n}{n}-\frac{1}{n}\ln\left(1+b_ne^{-(b-a)n}-a_nb_ne^{-bn}\right)\\ \stackrel{a_n,b_n=\mathcal O(n^p)}{=}&a+o(1)\hspace{40ex}\Box\end{aligned}$
$\bullet$ If $a=0$ or $b=0$ the quantity equals $0$.
$\bullet$ For $a,b\neq0$ and $a\neq1$, from Taylor's theorem we can write
$\sum_{k=0}^{n}\frac{(an)^k}{k!}=e^{an}-\frac{1}{n!}\int_{0}^{1}e^t(an-t)^n\,dt:=e^{an}-A_n,$
and
$A_n\stackrel{t=any}{=}\frac{(an)^{n+1}}{n!}\int_{0}^{1}e^{any}(1-y)^n\,dy=\frac{(an)^{n+1}}{n!}\mathcal O(1),$
so
$\begin{aligned}e^{-an}\sum_{k=0}^{n}\frac{(an)^k}{k!}&=e^{-an}\left(e^{an}-\frac{(an)^{n+1}}{n!}\mathcal O(1)\right)\\&\stackrel{(1)}{=}1-e^{(1-a+\ln a)n}\mathcal O(\sqrt{n})\mathcal O(1)\\&=1-e^{(1-a+\ln a)n}\mathcal O(\sqrt{n})\end{aligned}$
Now the given quantity equals
$-\frac{1}{n}\ln\left(1+\left(e^{-abn}-1\right)\left(1-e^{(1-a+\ln a)n}\mathcal O(\sqrt{n})\right)\right)$
and since $1-a+\ln a<0$ from Claim 1 the quantity goes to $\min\{a-1-\ln a,ab\}$.
$(1)$ From Stirling's formula
$\bullet$ For $a=1$ and $b\neq0$, proceeding as above, we need a better estimate for $\int_{0}^{1}e^{ny}(1-y)^n\,dy$ than just $\mathcal O(1)$.
Claim 2 $\hspace{5ex}$ For $a>-1$ and $c>0$ it is $\int_{0}^{c}t^ae^{-nt}\,dt = \frac{\Gamma(a+1)}{n^{a+1}}+\mathcal O(n^{-1}e^{-nc}).$
Indeed $\hspace{3ex}$ for $a>-1$ and $c>0$ :
$\begin{align}\int_{0}^{c}t^ae^{-nt}\,dt &= \int_{0}^{+\infty}t^ae^{-nt}\,dt-\int_{c}^{+\infty}t^ae^{-nt}\,dt\notag\\ &\stackrel{nt=x}{=} \frac{\Gamma(a+1)}{n^{a+1}}-\frac{1}{n^{a+1}}\int_{nc}^{+\infty}x^ae^{-x}\,dx\notag\\ &\stackrel{x=(1+u)nc}{=} \frac{\Gamma(a+1)}{n^{a+1}}-c^{a+1}\int_{0}^{+\infty}(1+u)^ae^{-(1+u)nc}\,du\notag\\ &\stackrel{1+u\leq e^u}{\Rightarrow} \notag\\ \left|\int_{0}^{c}t^ae^{-nt}\,dt-\frac{\Gamma(a+1)}{n^{a+1}}\right| &\leq c^{a+1}e^{-nc}\int_{0}^{+\infty}e^{u(a-nc)}\,dt=\mathcal O(n^{-1}e^{-nc})\notag\\ &\Longrightarrow \notag\\ \int_{0}^{c}t^ae^{-nt}\,dt &= \frac{\Gamma(a+1)}{n^{a+1}}+\mathcal O(n^{-1}e^{-nc}).\hspace{25ex}\Box\notag \end{align}$
Now looking at
$\int_{0}^{1}e^{nx}(1-x)^n\,dx$
we set $\phi(x)=-x-\ln(1-x)$ so
$\frac{1}{\phi^{'}(x)}=-1+\frac{1}{x}$
but
$t:=\phi(x)=\frac{x^2}{2}+\mathcal O(x^3)=\frac{x^2}{2}\left(1+\frac{2x}{3}+\mathcal O(x^2)\right)$
so, as $t\geq0$, setting $u=\sqrt{2t}$ we get
$u=x\left(1+\frac{2x}{3}+\mathcal O(x^2)\right)^{1/2}=x+\frac{x^2}{3}+\mathcal O(x^3)$
and reverse setting $x=u+a_2u^2+\mathcal O(u^3)$ and equating coefficients to get
$x=u-\frac{1}{3}u^2+\mathcal O(u^3).$
This will give
$\frac{1}{\phi^{'}(x)}=\frac{1}{\phi^{'}\left(\phi^{-1}(t)\right)}=\frac{\sqrt{2}}{2}t^{-1/2}-\frac{2}{3}+\mathcal O(t^{1/2})$
and applying Claim 2
$\int_{0}^{1}e^{ny}(1-y)^n\,dy=\frac{\Gamma(1/2)}{\sqrt{2}}n^{-1/2}+\mathcal O(n^{-1})=\sqrt{\frac{\pi}{2}}n^{-1/2}+\mathcal O(n^{-1}).$
With Stirling's formula again
$\begin{aligned}e^{-n}\sum_{k=0}^{n}\frac{n^k}{k!}&=1-\left(\frac{1}{\sqrt{2\pi}}n^{1/2}+\mathcal O(n^{-1/2})\right)\left(\sqrt{\frac{\pi}{2}}n^{-1/2}+\mathcal O(n^{-1})\right)\\&=\frac{1}{2}+\mathcal O(n^{-1/2})\end{aligned}$
so the given quantity equals $-\frac{1}{n}\cdot\mathcal O(1)\to0$.