Yes.
To build a minimum spanning tree you will order edges by weights from smallest to largest. An edge is included as long as it does not make a cycle. If there are two minimum weight edges, then these edges are first and second. The first two edges do not make a cycle. Therefore, since all other edges are distinct, then you will end up with one minimum spanning tree. That is, whatever order of selecing the two minimum edges, the tree would be the same. QED.
A bit of clarification:
Whether you are using Prim or Kruskal, the same result applies. Here is why. Kruskal order ALL edges ascendingly. In Prim, at each node $v$, the edges are ordered ascendingly in this manner $(v,u)$ where $u$ is a neighbor of $v$. Whenever an edge $(u,v)$ is selected, the edge $(v,u)$ is deleted [note they are the same since it is an undirected edge]. The edge with smallest minimum weight $e _{min(1)} =(x,y)$ is the first in the global ordering. It also should be the first at node x and y otherwise it is not the smallest. The argument applies to the second minimum edge $e _{min(2)}$.
In general, Kruskal follows a total order of edges, while Prim uses a partial order of edges.