Your problem is equivalent to End-of-Metered-Line.
This can be shown by reducing your problem to End-of-Potential-Line (see https://arxiv.org/abs/1702.06017). This is a version of End-of-the-Line where the line is equipped with a potential function, and a solution is either the end of a line, or a vertex on the line at which the potential function decreases. This differs from EOML, since the potential is not required to increase by exactly one at each vertex.
We can reduce your problem to EOPL by taking the successor of each vertex to be the neighbor with the larger index, and the predecessor of each vertex to be the neighbor with the smaller index. Any vertex that has two neighbors with smaller indices is the end of a line, and any vertex that has two larger neighbors is also the end of a line. The potential function of a vertex is just its index, which is monotonically increasing along each line.
It was shown in https://arxiv.org/abs/1702.06017 that EOPL is equivalent to EOML, under polynomial-time reductions, and thus is in CLS.
For the other direction, EOML can be reduced to your problem by embedding the potential space into the vertex space using the higher-order bits. In other words, each vertex $v$ in the EOML instance is mapped to the integer $K \cdot p(v) + v$, where $p(v)$ is the potential of $v$, and $K$ is some constant that is larger than the total number of vertices in the EOML instance.
So to answer your question, the problem is unlikely to be PPAD-complete since that would imply PPAD $\subseteq$ CLS.