$\newcommand{\lln}{\operatorname{\lln}}\newcommand{\bb}{\binom}$We are going to deduce from \eqref{1} a bound better than the ultimate bound in \eqref{2}. Actually, we are going to obtain the exact asymptotic of the upper bound in \eqref{1}. 

(In particular, it follows that the ultimate bound in \eqref{2} is not optimal. Already for this reason, it seems hard, if at all possible, to reconstruct the logic of Erdős and Rényi that allowed them to get \eqref{2} from \eqref{1}.) 

To begin our consideration, note that the upper bound on $P$ in \eqref{1} is 
\begin{equation*}
	p_n:=\sum_{s=2}^{\ln\ln n}\bb ns \sum_{r=1}^{\bb s2} \bb{\bb s2}r f_{s,r}, 
\end{equation*}
where 
\begin{equation*}
	f_{s,r}:=\frac{\dbinom{\bb{n-s}2}{N_c-r}} {\dbinom{\bb n2} {N_c}}. 
\end{equation*}

Note that 
\begin{equation*}
	f_{s,r}=\frac{P_1 P_2}{P_3},
\end{equation*}
where 
\begin{equation*}
\begin{aligned}
	P_1&:=N_c\cdots(N_c-r+1), \\ 
	P_2&:=\Big(\bb n2-N_c\Big)\cdots\Big(\bb{n-s}2-N_c+r+1\Big), \\ 
	P_3&:=\bb n2\cdots\Big(\bb{n-s}2+1\Big).  
\end{aligned}
\end{equation*}
Here and in what follows, $s$ is an integer in the interval $[2,\ln\ln n]$ and $r$ is an integer in the interval $[1,\bb s2]$. 

To determine the asymptotics of $P_1,P_2,P_3$, we are going to use the following simple lemma, which will be proved at the end of this answer. 

**Lemma 1:** If $a$ and $b$ are positive integers varying so that $b^3=o(a^2)$, then 
\begin{equation*}
	(a+b-1)\cdots a\sim\Big(a+\frac{b-1}2\Big)^b. 
\end{equation*}

Using Lemma 1 and letting 
\begin{equation*}
	q_{n,s}:=\bb n2-\bb{n-s}2=s(n-(s+1)/2), 
\end{equation*}
uniformly in $s,r$ as specified above we get the following (as $n\to\infty$): 
\begin{equation*}
	P_1\sim\Big(N_c-\frac{r-1}2\Big)^r
	=N_c^r\Big(1-\frac{r-1}{2N_c}\Big)^r\sim N_c^r,
\end{equation*}
\begin{equation*}
	P_3\sim\Big(\bb n2-\frac{q_{n,s}-1}2\Big)^{q_{n,s}}
	=\bb n2^{q_{n,s}}\Big(1-\frac{q_{n,s}-1}{n(n-1)}\Big)^{q_{n,s}} \\ 
	\sim \bb n2^{q_{n,s}} e^{-s^2}, 
\end{equation*} 
\begin{equation*}
\begin{aligned}
	P_2&\sim\Big(\bb n2-N_c-\frac{q_{n,s}-r-1}2\Big)^{q_{n,s}-r} \\ 
	&=\bb n2^{q_{n,s}-r}\Big(1-\frac{2N_c+q_{n,s}-r-1}{n(n-1)}\Big)^{q_{n,s}-r} \\ 
	&\sim \bb n2^{q_{n,s}-r} e^{-s\ln n-2cs-s^2}.  
\end{aligned}
\end{equation*} 
So, 
\begin{equation*}
	f_{s,r}=\frac{P_1 P_2}{P_3}
	\sim\rho_n^r e^{-2cs}n^{-s},
\end{equation*}
where 
\begin{equation*}
	\rho_n:=\frac{N_c}{\bb n2}. 
\end{equation*}
So, 
\begin{equation*}
\sum_{r=1}^{\bb s2} \bb{\bb s2}r f_{s,r}
=\Big((1+\rho_n)^{\bb s2}-1\Big)e^{-2cs}n^{-s}\sim\rho_n\bb s2 e^{-2cs}n^{-s} \\ 
\sim\bb s2\frac{\ln n}n\,e^{-2cs}n^{-s}. 	
\end{equation*}
So, 
\begin{equation*}
\begin{aligned}
	p_n&\sim\frac{\ln n}n\,\sum_{s=2}^{\ln\ln n}\bb ns \bb s2 e^{-2cs}n^{-s} \\ 
	&\le\frac{\ln n}n\,\sum_{s=2}^\infty\frac1{s!} \bb s2 e^{-2cs} \\ 
	&=C\frac{\ln n}{n}, 
\end{aligned}
\end{equation*}
where $C:=\frac{1}{2} e^{e^{-2 c}-4 c}$. 

Moreover, by dominated convergence,
\begin{equation*}
	p_n\sim C\frac{\ln n}{n}.  \quad\Box  
\end{equation*}

----

It remains to present  

*Proof of Lemma 1:* 
\begin{equation*}	
\begin{aligned}
	\big((a+b-1)\cdots a\big)^2&=\prod_{j=0}^{b-1}(a+j)(a+b-1-j) \\ 
&=\prod_{j=0}^{b-1}\Big[\Big(a+\frac{b-1}2\Big)^2-\Big(\frac{b-1}2-j\Big)^2\Big] \\  
&=\Big(a+\frac{b-1}2\Big)^{2b}
\prod_{j=0}^{b-1}\Big[1-\frac{(b-1-2j)^2}{(2a+b-1)^2}\Big] \\   
&=\Big(a+\frac{b-1}2\Big)^{2b}
\prod_{j=0}^{b-1}\Big[1-O\Big(\frac{b^2}{a^2}\Big)\Big] \\  
&=\Big(a+\frac{b-1}2\Big)^{2b}
\Big[1-O\Big(\frac{b^3}{a^2}\Big)\Big] \\ 
&\sim \Big(a+\frac{b-1}2\Big)^{2b}. \quad\Box 
\end{aligned}
\end{equation*}