I have a new result that might be a step towards proving there are infinitely many solutions.  We show that there is an infinite sequence of values for which $F(s)$ is at most 1.

Consider $s_k = 2^{2^k} - 5$ for $k \ge 3$.

We have stated that a number is in $S$ if it is not $4 + p$ for prime $p$, and is not $3q$ for prime $q$.  We also stated that a product is in $P$ iff exactly one divisor pair has sum in $S$.  We will show that $s_k \in S$ and at most one of the products $l (s_k - l)$ is in $P$.

$s_k$ is not a multiple of 3, so the latter condition for inclusion in $S$ is always satisfied.

Subtract 4 to get $2^{2^k} - 9$.  This number is not prime.  It is divisible by 7 is $k$ is even, and 13 if $k$ is odd.  This can be proven by induction.  Therefore $s_k \in S$ for all $k \ge 3$.

Consider ways to sum to $s_k$.  Since $s_k$ is odd, all pairs of summands must consist of one even and one odd number.  Let the even number be $a'$, the odd number $b$.  Let $v_p(x)$ be the p-adic valuation of $x$.  Define $m = v_2(a')$ and $a = 2^{-m} a'$.  We know $m \in [1, 2^k)$ and $a$ is odd.  We can write $s_k = 2^m a + b$.

If $a \ge 3$, write $s'_k = 2^m b + a$.

Add $s_k$ and subtract it:

$s'_k = 2^m b + a + (2^m a + b) - (2^{2^k} - 5)$

Factor and subtract 4:

$s'_k - 4 = (2^m + 1)(a + b) - (2^{2^k} - 1)$

Since $m < 2^k$, we know $2^m + 1$ shares a nontrivial factor with $2^{2^k} - 1$.  This factor divides $s'_k - 4$, and thus $s'_k - 4$ is composite.  Furthermore, $s'_k$ is congruent mod 3 to $(-1)^m s_k$, so it is not a multiple of 3.  $s'_k \in S$, so the product $2^m a b$ with $a \ge 3$ has at least two factorizations with sum in $S$, and cannot be in $P$.

If $a = 1$, we have $b = 2^{2^k} - 5 - 2^m$.  For $m \ge 3$, we can examine $b$ modulo $y = 2^{2^{v_2(m-2)}} + 1$.

Write $2^m$ as $4 * 2^{m-2}$, and then write $2^{m-2}$ as $2^{2^{v_2(m-2)} m'}$ with $m'$ odd.  Modulo $y$, $2^m = 4(-1)^{m'} = -4$.
Since $2^k > m$, we know $2^{2^k}$ is $2^{2^{v_2(m-2)}}$ raised to some even power.  Thus, modulo $y$, $b = 1 - 5 + 4 = 0$.  We found a divisor of $b$.

When $m$ is even, the alternate factor sum $y + 2^m (b/y)$ is 1 mod 6, and thus is in $S$, so $2^m b \not \in P$.

When $m$ is odd, $y = 3$, so we know $b$ is a multiple of 3.  Define $r = v_3(b)$, $z = 3^{-r} b$.  We know $r$ is at least 1 and $z$ is congruent to either 1 or 5 mod 6. If $z \ge 5$, we can write the factor sums $s'' = z + 3^r * 2^m$ and $s'''= 3^r + 2^m z$.

If $z$ is 1 mod 6, $s''$ is 1 mod 6.  If $z$ is 5 mod 6, $s'''$ is 1 mod 6.  Either way, one of the two values is in $S$, so $2^m 3^r z \not \in P$.

$z = 1$ occurs when there is a solution to the diophantine equation $2^{2^k} - 5 = 2^m + 3^n$.

I am to believe that there is exactly one solution for $k \ge 3$.  When $k = 3$, we have $2^8 - 5 = 251 = 8 + 243$.  In this case, we can consider the alternate factor sum $27 + 72 = 99$, which is in $S$.  A result told to me [here](https://math.stackexchange.com/questions/2811957/solutions-to-22k-5-2m3n?noredirect=1#comment5798340_2811957) claims there can be no other cases where $z = 1$ can occur.

We have eliminated $a \ge 3$ unconditionally and $a = 1$ when $m \ge 3$.  We are left with only one possible pair of summands: 4 and $2^{2^k} - 9$.  I don't have a proof that $4(2^{2^k} - 9)$ is or is not in $P$.  By my counts it certainly can be, if 3 derived numbers are all prime.  That is a rare event, and it would be reasonable for it never to occur given how quickly $2^{2^k}$ grows.  However, our initial claim was that at most one derived product is in $P$.  We have shown that.

This completes the proof.