We prove the following.
Claim. The set $S$ is given explicitly by $S = \{x(x+1)y(y+1) \mid x, y > 0\}$.
Proof. Note that $m, m+1 \in \mathcal{F}(n)$ if and only if there exist integers $q > p \ge 0$ such that
$$m^2 - 4n = p^2 \\ (m+1)^2 - 4n = q^2.$$
Subtraction gives $2m + 1 = q^2 - p^2$, so $p$ and $q$ are different parities. We can thus let $q - p = 2x + 1$, $q + p = 2y + 1$, where $y \ge x \ge 0$ are integers. It follows that
$$\begin{aligned} 4n &= m^2 - p^2 \\ &= \left( \frac{q^2 - p^2 - 1}{2} \right)^2 - p^2 \\ &= \left( \frac{q^2 - p^2 - 1}{2} - p \right) \left( \frac{q^2 - p^2 - 1}{2} + p \right) \\ &= \frac{q^2 - (p^2 + 2p + 1)}{2} \cdot \frac{q^2 - (p^2 - 2p + 1)}{2} \\ &= \frac{1}{4}(q - p - 1)(q - p + 1)(q + p - 1)(q + p + 1) \\ &= \frac{1}{4}(2x)(2x + 2)(2y)(2y + 2) \\ \implies n &= x(x + 1)y(y + 1). \end{aligned}$$
Since $n > 0$ we require $x, y > 0$. Conversely, if $n = x(x+1)y(y+1)$ for positive $x$ and $y$ then $m = \sqrt{p^2 + 4n} = \sqrt{(y-x)^2 + 4n} = 2xy + x + y = x(y+1) + (x+1)y$ and $m + 1 = 2xy + x + y + 1 = xy + (x+1)(y+1)$. Thus we conclude the main claim. $\square$

From this, part (a) follows as
$$\sum_{n \in S} n^{-1} \le \left( \sum_{x \ge 1} \frac{1}{x(x+1)} \right) \left( \sum_{y \ge 1} \frac{1}{y(y+1)} \right) = 1 \cdot 1 = 1.$$

As for (b), retain the notation in the proof of the claim. Now $m + 2 \in S$ if and only if $(m+2)^2 - 4n$ is a square, say $r^2$. Writing in terms of $p$ and $q$ as parameters we find
$$\begin{aligned} r^2 &= (m+2)^2 - 4n = m^2 - 4n + 4m + 4 = p^2 + 2 + 2(2m+1) \\ &= p^2 + 2(q^2 - p^2) + 2 = 2q^2 - p^2 + 2 \\ \iff 2q^2 + 2 &= p^2 + r^2 \quad (\dagger) \end{aligned}$$

with $q > p$ of different parity and $n = \frac{1}{16}(q-p-1)(q-p+1)(q+p-1)(q+p+1)$.
Note that (by taking modulo 8) we have $q \not\equiv p \equiv r \pmod 2$, and so there are no parity issues and we will always assume $p < q < r$ in $(\dagger)$. Now, for every $q$, the equation $(\dagger)$ has a canonical solution $(p, r) = (q-1, q+1)$, but this leaves $n = 0$. Thus we want to show for infinitely many $q$ there is a third way to write $2q^2 + 2$ as a sum of squares, which will give the desired $p$.
To do this, choose large integers $q$ such that $q^2 + 1$ is divisible by at least three distinct $1 \bmod 4$ primes. Since each such prime can be written as a sum of two squares, using Lagrange identity, we can deduce that $2q^2 + 2$ can be written as a sum of two squares in at least three different ways, as desired.