We look for an appropriate $S$. The smallest value in $\{4,...,100\}$ that is in $T$ is $11$. This can be seen as follows:
$$
2\cdot 2\in R \text{ and } 2\cdot 2\in C_{4} \text{, Therefore } 4\notin T \\
3\cdot 2\in R \text{ and } 3\cdot 2\in C_{5} \text{, Therefore } 5\notin T \\
3\cdot 3\in R \text{ and } 3\cdot 3\in C_{6} \text{, Therefore } 6\notin T \\
5\cdot 2\in R \text{ and } 5\cdot 2\in C_{7} \text{, Therefore } 7\notin T \\
5\cdot 3\in R \text{ and } 5\cdot 3\in C_{8} \text{, Therefore } 8\notin T \\
7\cdot 2\in R \text{ and } 7\cdot 2\in C_{9} \text{, Therefore } 9\notin T \\
7\cdot 3\in R \text{ and } 7\cdot 3\in C_{10} \text{, Therefore } 10\notin T
$$
While:
$$
A_{11} = \{(2,9), (3,8), (4,7), (5,6)\} \text{, and so }\,\,C_{11} = \{18, 24, 28, 30\}
$$
We proceed to check if there if there exists a unique $P\in C_{11}$, for which $|D_P \cap T|=1$. Looking at $D_{18}=\{9,11\}$, and $D_{24}=\{10,11,13\}$, and noting that $11\cdot 2\in R$ and $11\cdot 2\in C_{13}$, Therefore $13\notin T$, we see:
$$
|D_{18}\cap T| = |\{11\}| = 1\,\text{ and }\,|D_{24}\cap T| = |\{11\}| = 1
$$
So we have more than one value that satisfies the statement. However we required that the value be unique, so we must conclude that $S\neq 11$. The next value in $T$ after $11$ is $17$. This can as can be seen as follows:
$$
7\cdot 5\in R \text{ and } 7\cdot 5\in C_{12} \text{, Therefore } 12\notin T \\
7\cdot 7\in R \text{ and } 7\cdot 7\in C_{14} \text{, Therefore } 14\notin T \\
13\cdot 2\in R \text{ and } 13\cdot 2\in C_{15} \text{, Therefore } 15\notin T \\
13\cdot 3\in R \text{ and } 13\cdot 3\in C_{16} \text{, Therefore } 16\notin T
$$
While:
$$
A_{17} = \{(2,15),(3,14),(4,13),(5,12),(6,11),(7,10),(8,9)\},\,C_{17} = \{30, 42, 52, 60, 66, 70, 72\}
$$
As before, we proceed to check if there if there exists a unique $P\in C_{17}$, for which $|D_P \cap T|=1$. Let us first write out the sets $D_i$ for all $i\in C_{17}$:
$$
\begin{align}
&D_{30} = \{11, 13, 17 \} \\
&D_{42} = \{13, 17, 23 \} \\
&D_{52} = \{17, 28 \} \\
&D_{60} = \{16, 17, 19, 23, 32 \} \\
&D_{66} = \{17, 25, 35 \} \\
&D_{70} = \{17, 19, 37 \} \\
&D_{72} = \{17, 18, 22, 27, 38 \} \\
\end{align}
$$
Next, we note the following: A quick way to check if an odd number $k$ can be written as the sum of two primes is to check if $k-2$ is prime. This is because the primes in the sum must have different parity, and the only even prime is $2$. By definition, if $k$ cannot be written as $(1)$ the sum of two primes, nor as $(2)$ $p^2 + p$ for some prime $p$, then $k\in T$. The first few values for $p^2 + p$ are:
$$
\begin{array}{c|c}
p & p^2+p \\
\hline
2 & 6 \\
3 & 12 \\
5 & 30 \\
7 & 56 \\
\end{array}
$$
Using the data in this table along with our observation regarding the way odd itegers must be written as the sum of primes, we conclude that $23, 27, 35, 37 \in T$, and therefore:
$$
|D_i \cap T| > 1
$$
for $i \in \{30,42,60,66,70,72\}$. Finally, we see that $23\cdot 5\in R$ and $23\cdot 5\in C_{28}$, Therefore $28\notin T$, so that:
$$
|D_{52}\cap T| = |\{17\}| = 1
$$
Therefore our unique product is $P = 52$, and the answer to the puzzle is $4$ and $13$.