a cut containing exactly one edge in each path
It really depends on what do you exactly mean by "path".

Edge-disjoint paths -- Given a graph $G=(V,E)$, a minimum edge-cut consists of one edge for every edge-disjoint path between $s$ and $t$.

Vertex-disjoint paths -- Given a graph $G=(V,E)$, a minimum vertex-cut consists of one vertex for every vertex-disjoint path between $s$ and $t$. Therefore, the set of edges incident to the vertices in the cut (out-edges) consist of one edge per vertex-disjoint path between $s$ and $t$.

Having exactly one edge per every simple path between $s$ and $t$ is impossible. The reason is that the number of edges in the cut must be equal to the number of simple paths between $s$ and $t$, however, the number of simple paths between two vertices $s$ and $t$ can be exponential in terms of $n$, while the number of edges is at most $n^2$. Using Pigeonhole principle, it can be shown that it is impossible to find such solution.

The following example shows that the number of paths between $s$ and $t$ is $2^{(n-2)/2}$, while the number of edges between them is $2n-4$.



If you are interested in finding a set of edges that consists of at least one edge from every path between $s$ and $t$, then you can use the minimum edge-cut between $s$ and $t$.
If you are considering a case where you need at least one edge from each path, and no more than one edges per path, you can use an approach that is similar to minimum cut algorithm, but I am not sure how efficient it would be in practice.


Find the minimum edge-cut between $s$ and $t$;
Using a graph traversal algorithm (i.e. DFS), find if there is a path that goes over two of the edges in the cut;
If you find a such a path, remove those edges $(u_1, v_1), \cdots, (u_k, v_k)$ from the cut $C$ and replace them with the minimum cut from each $u_i$ to $t$ ($1 \leq i \leq k$), and repeat from 2;
If there is no such path then output $C$ and terminate;