The complexity of a multi-objective shortest path problem
It is NP-hard by reduction from Partition.

Let $x_1,\dots,x_n$ be the partition instance.

Create a graph on levels $0,1,\dots,n$ with level $0$ containing the start vertex $s$, and level $i$ containing two vertices, $a_i$ and $b_i$.

Every path from $s$ to a vertex in the last level corresponds to a partition of $x_1,\dots,x_n$ into two blocks, $A$ and $B$, as follows:
If the path goes through $a_i$, then $x_i$ is placed in $A$, otherwise the path must go through $b_i$, and $x_i$ is placed in $B$.
Conversely, every partition is realized by some path in this way.

For every edge to $a_i$ from the level below, let the cost be $(x_i,0)$, and for every edge to $b_i$ from the level below, set the cost to $(0,x_i)$.
Finally, set $L_i = \sum_{i=1}^n x_i/2$.
Now the partition instance has a balanced partition if and only if there is a solution to the path problem with cost $\le \sum_{i=1}^n x_i/2$.

(Edit: I originally gave a reduction from the shortest weight-constrained path problem, and then to answer a comment; a reduction from partition to the shortest weight-constrained path problem. However, the last reduction is enough by itself.)