$\newcommand{\de}{\delta}$
The bound ($\star\star$) that you want is impossible in general. E.g., take $A_n=0$, $C_n=1$, $B_n=n\Delta t$ for all $n$, with $\Delta t>0$. Then ($\star$) will hold, whereas ($\star\star$) will not hold for large enough $n$, for any choice of $f,g,h$.
Added: The bound ($\star\star$) will hold with $f=g=1$ and $h=b:=\frac{K+1}p$ if we additionally assume that $A_n$ dominates $B_n$ in the sense that $B_n\le KA_n$ for some real $K>0$ and all $n$. Indeed, let $\de:=\Delta t$, $S_n:=A_n+B_n$, and $M_n:=\max_{0\le k \le n}C_k$. Then, with such $f,g,h$, ($\star\star$) can be rewritten as \begin{equation} S_n\le S_0+bM_n. \tag{!!} \end{equation} On the other hand, the condition $B_n\le KA_n$ can be rewritten as $(1+p\de)A_n + B_n\ge(1+\de/b)S_n$; so, ($\star$) yields \begin{equation} (1+\de/b)S_n\le S_{n-1}+C_n\de. \end{equation} Now it is easy to to prove (!!) by induction. Indeed, for $n=0$ (!!) is trivial. Assuming (!!) holds with $n-1$ in place of $n$, we have \begin{multline} (1+\de/b)S_n\le S_{n-1}+C_n\de\le S_0+bM_{n-1}+C_n\de \le(1+\de/b)S_0+(b+\de)M_n \\ =(1+\de/b)(S_0+bM_n), \end{multline} so that (!!) indeed follows.