Minimal paths as solution of a linear program of a special network flow
There is a major flaw in your question: You presented a shortest path formulation and not a special case of a network flow problem.
Your objective function minimizes the cost of the path. The next 5 constraints are the classic flow constraints, which ensure a path between $s$ and $t$. The last constraint is just the bound of variable $x$.

Now, we can start answering your questions: 




  Given that every s−t-path fulfills the constraints, we can conclude that if S is an s−t-path, then it is an s−t-path with minimal costs.


Correct. It is the shortest path on the defined graph.




  Further, S can't contain any cycle with positive weight, as we could simply remove this cycle from S and end up with a lower-cost solution that fulfills the constraints.


The solution of a shortest path problem does not contain a cycle, for obvious reasons.




  Did I miss something in the proof?


Sorry, but your proof does not show anything. You are showing that the shortest path is the shortest path.




  This whole procedure seems like a lot for a less efficient method to obtain a shortest path


Because you are solving the shortest path problem :)
However, I still recommend you to stick with Dijkstra's algorithm, which is faster than linear programming.




  Besides the non-negativity constraint, all constraints are actually equalities.


Yes, good observation. You just have the classic flow constraints (which are all equalities). I recommend you to read the Ahuja's book on network flows (is the best I know, by the way).




  The algorithm should be easily adaptable to also allow for negative cycles (by putting some linear restriction on the solution, e.g. that the path mustn't be too long, etc.)


You can do that. However, this constraint lies in enumerating all the cycles of your graph, which cannot be fastly solved.