Why does Fleury's algorithm have to return to the starting node?
Suppose you end up at a vertex different from the start. You may have visited the vertex before, say $k$ times. Each time you visited the vertex you used up one edge to enter it and one edge to leave it. So $2k$ edges, an even number. When you enter it for the last time, you used up $2k+1$ edges, an odd number. Since the degree of all nodes is even, there must be an edge left over which you can leave the vertex and the algorithm wouldn't have stopped.

The start vertex is special because you don't need to enter it to visit it the first time (right at the start).

Note that a graph can also be eulerian if it has exactly two vertices with odd degree. Then you have to choose one of them as a start vertex and you will end up at the other odd-degree vertex at the end of the algorithm. I leave it to you to prove why this is the case.