Formally one can define: $ \begin{align} &A_x=\{(a,b) \in\mathbb{N}^2 : a+b=x,\,\,1 Then the problem becomes finding the unique value $S\in\{4,...,100\}$, such that the following statement holds: $ S\in T\,\text{ and }\,\exists!P\in C_S\,:|D_P \cap T|=1 $ Then $S$ is our sum, and the unique value $P$ in the above statement is our product. With some effort we can find these values, by simple brute force search.
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 $13$.