If $s$ is not too large, the inequality is correct. This is because the quotient of binomial coefficients is $$ \prod_{0\le i$N_c=\frac{1}{2}n\log n+cn-\theta$, for some $0\le\theta<1$, and inserting this into this last inequality cancels the terms $s\log n$ and $-2cs$, leaving $$ \frac{2s(n-s)}{n^2}\theta +\frac{2s^2}{n^2}(\frac{1}{2}n\log n+cn)\le 3s, $$ which is true if we require that $s\le n/\log n$ and that $n$ be sufficiently large.
If $s=\Omega(n)$, the inequality will not hold. Set $s:=\lfloor n\delta \rfloor$, fix $\delta\in(0,\frac{1}{2}]$ and $c$, and let $n$ become large. Look at the logarithms of each side. The quotient of binomial coefficients on the left-hand side will give $$N_c \log(1-s(n-s){n\choose 2}^{-1})+O((\log n)^2)=\frac{1}{2}(n\log n) \log(1-2\delta(1-\delta)) + O(n),$$ while $\log {n\choose s}=O(n)$. On the right-hand side, the logarithm of $1/s!$ is $-\delta n \log n + O(n)$, and $\log e^{(3-2c)s}$ is $O(n)$. Therefore, if the inequality were true, we would have $$ \frac{1}{2}( n\log n) \log(1-2\delta(1-\delta))+O(n)\le -\delta n\log n+O(n), $$ but since $\frac{1}{2}\log (1-2\delta(1-\delta))>-\delta$ for all $\delta$ in $(0,\frac{1}{2}]$, this is false for large enough $n$.
The failure of the inequality for large $s$ is not a problem for the Erdős-Rényi paper. This is because the use of this inequality (numbered (14) in the paper) is to bound the sum of the left-hand side of (14) as $s$ varies between some lower bound and $n/2$. However, the quotient of binomial coefficients is decreasing over $0\le s\le n/2$, so if we set $s_0:=n/\log n$ and sum the left-hand side of (14) over $s_0\le s\le n/2$, the result will be no more than $$ 2^n \left.\binom{\binom{n}{2}-\lceil s_0\rceil(n-\lceil s_0\rceil)}{N_c}\right/\binom{\binom{n}{2}}{N_c} \le 2^n \exp -\frac{2s_0(n-s_0)}{n^2} N_c, $$ and taking the logarithm of the right-hand side gives $$ n\log 2 - \frac{2s_0(n-s_0)}{n^2} (\frac{1}{2}n\log n+cn-\theta). $$ However, if we fix $c$, this is $-n(1-\log 2)+O(n/\log n)$, which approaches $-\infty$ as $n$ becomes large. Therefore, the right-hand side of the inequality (13) in the paper may be split into four pieces: (1) $M$s$ is replaced by $n-s$.