You know that a graph is bipartite if and only if it contains only cycles of even length. If (at least) one $s\in S$ contains a cycle of odd length, you know $G$ is not bipartite. If for all $s\in S$, $s$ contains only even cycles, you want $G$ to be bipartite. Suppose on the contrary that $G$ does have an odd cycle. If it does not visit al $v\in V$, you can select a $v'$ not in the cycle, and look at $G-v'$. This contains a odd cycle, so we have a contradiction. So we know that $G$ can only contain odd cycles of odd length which visit all vertices. If $|G|$ is even, we're done. when the only odd cycle visits all vertices, we have a problem. If there is at least one other edge, we know that is devides the long odd cycle in two halfs, one of which has even length, and together with the extra edge, we have a shorter cycle of odd length, so contradiction. The only case where we cannot (see edit) determine whether $G$ is bipartite or not is when $G$ is a cycle of odd length. This can easily be seen when looking at $G=K_3$.
EDIT In fact, it is possible in another way: Count the total number of edges in all thesubgraphs. If this is equal to $n(n-2)$, where $n$ is the size of $G$ and $n-1$ the size of the subgraphs, $G$ must have had $n$ edges. If it had at most $n-1$ edges, you would not count $n(n-2)$ edges, because $n(n-2)$ is the maximum, and it is only reached when $G$ had $n$ edges.