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.
