Bridges and Edge Disjoint Paths
Yes, it is true that there exist two edge-disjoint paths between any two vertices in a connected bridgeless graph.

Define a closed trail as a trail whose last vertex is its first vertex. Note, a trail contains two edge-disjoint paths between any two of its vertices.

Trail combining lemma. Let $G$ be a graph.  Suppose a closed trail passes vertex $v$. Another closed trail passes vertex $u$. If those two trails share one vertex, there is a closed trail that passes both $u$ and $v$.

Proof. You can convince yourself easily by drawing a few situations. Done. 



Let us prove the following proposition.

A graph is connect and bridgeless if and only if any two vertices are included in a closed trail.

Proof.

"$\Leftarrow$": Obvious.

"$\Rightarrow$": Suppose $G$ is connected and bridgeless. $v$ and $w$ are two vertices. 

Let $$T_v=\{u : u \text{ is a vertex of } G \text{ such that there is closed trail that passes }v\text{ and } u\}.$$ 
Suppose $u\in T_v$ and $s$ be a neighboring vertex of $u$, i.e., $\{u,s\}$ is an edge. 


There is a closed trail that passes $v$ and $u$ by the definition of $T_v$.
There is a closed trail that passes $u$ and $s$, since edge $\{u,s\}$ is not a bridge. 


Thanks to the lemma, we have a closed trail that passes $v$ and $s$. In other words, $T_v$ contains all neighborhoods of all its vertices. That means $T_v$ (with all edges between its points) is a connected component, which must be $G$ itself. In particular, there is a closed trail that passes $v$ and $w$. Proof is complete.



A very similar characterization of bridgeless connected graphs is Robbins' theorem. If $G$ is connected and bridgeless, that theorem says we can choose a direction for each edge of $G$, turning it into a directed graph that has a path from every vertex to every other vertex.