We shall use graph theory to answer this question. Let $G$ be an undirected graph. Assume a vertex for each compound. Also there is an edge between two vertices if and only if the two compounds react. As an example case suppose that all the elements in $S$ react with only elements in $S$. So in the induced subgraph corresponding the vertices in $S$, each vertex has odd degree and there are $9$ vertices in $S$. So we have odd no. of vertices of odd degree which is not possible. So the tightest example which we can have is as shown:
If the situation is as given and let us assume that the vertices in $U\backslash S$ which are not shown ($11$ such vertices are there) do not react with anything, then we reach at contradiction of :
Statement 1 : Because, here in the above example the $11 (= 23-9\text{(vertices of S)}-3\text{(vertices in blue)}$ vertices have even degree. So every vertices does not have odd degree.
Statement 3: Because, here in the above example, the $3$ blue vertices have odd degree [$1$ as shown]. So every vertex does not have even degree.
What about statement 2? In this example statement 2 is correct. Why? Let us assume that statement 2 is false. So all the $14$ vertices of $U\backslash S$ should have even degrees in graph $G$. But in $G-S$ the three blue vertices should have odd degrees, which is not possible, as a graph should have an even no of vertices with odd degrees.
But we cannot say about the correctness of a statement, just by proving it correct for one example case. We need to show that it is true for all cases. So again let us assume that statement 2 is false. So all the $14$ vertices of $U\backslash S$ should have even degrees in graph $G$. So their degree sum is also even. Also the $9$ vertices in $S$ have degree $3$ each. So sum their degree sum is $9* 3=27$ which is odd. So sum of degrees of the vertices of the graph = even+odd= an odd number. But degree sum of the vertices of a graph graph is twice the number of edges in the graph and hence even. So we arrive at a contradiction and statement 2 is true indeed. [This proof actually also proves that statement 3 is false].
A formal proof that statement 1 is wrong: if all $14$ vertices in $U\backslash S$ have odd degree then their degree sum shall be odd. Also we know degree sum of the vertices in $S$ is odd, so the degree sum of the vertices in $G$ shall be odd which is not possible,