Shortest path on a hypergraph with no leftovers
This answer doesn't answer the question about previous work, but it does show the problem is NP-complete.

Lemma 1. Finding a shortest $s$-$t$ hyperpath (as defined in the post) in a given hypergraph is NP-complete, even in hypergraphs of hyperedge degree 3.

Proof. Clearly the problem is in NP.  It is NP-hard by the following reduction from 3D-matching.

Fix an instance of 3D matching -- a set of triples $X$ from universe $U=\{1,2,\ldots,3n\}$.  Given the instance, the reduction outputs the instance of Shortest $s$-$t$ Hyperpath in the hypergraph $G=(V,E)$ defined as follows.

First create the vertex set $V=\{b_0, a_1, b_1, a_2, b_2, a_3, b_3, \ldots, a_{3n}, b_{3n}\}$.  Define the hyperpath source and destination to be $s=b_0$ and $t=b_{3n}$.  Next, add $3n$ "path" hyperedges $$(b_0,a_1,b_1), (b_1,a_2,b_2), (b_2, a_3, b_3), \ldots, (b_{3n-1}, a_{3n}, b_{3n}).$$  Finally, for each triple $(i, j, k)$ in $X$, add hyperedge $(a_i, a_j, a_k)$.  This completes the reduction.

We'll show that there is an $s$-$t$ hyperpath of length at most $4n$ if and only if there is a 3D matching.  

First suppose that there is a 3D matching $M$ (that is, an $n$-subset of $X$ such that each element in $\{1,\ldots,3n\}$ occurs in exactly one triple in $M$).  Then the $3n$ "path" hyperedges, together with the $n$ hyperedges $\{(a_i, a_j, a_k) : (i, j, k) \in M\}$ form the desired hyperpath.

Conversely, suppose there is such a hyperpath (from $b_0$ to $b_{3n}$).  By induction on the $b_i$ vertices, each of the $3n$ path hyperedges $(b_i, a_{i+1}, b_{i+1})$ must be in the hyperpath.  Then, since each vertex $a_1,\ldots,a_{3n}$ is in one such path hyperedge, each such vertex must also be in an odd number of other hyperedges in the hyperpath.  Those hyperedges must correspond to triples in $X$.  Each such hyperedge contains $3$ of the $a_i$ vertices, and the hyperpath has length $4n$ (or less), so there must be $n$ such hyperedges, and their corresponding triples must form a 3D matching. $~~\Box$

I suspect that even determining whether there exists any $s$-$t$ hyperpath is also NP-complete.

EDIT: I think the existence of an $s$-$t$ hyperpath can be determined in polynomial time by checking feasibility of this set of linear equations over $\mathbb{Z}_2$: introduce an indicator variable $x_e$ for each hyperedge $e$, then, for every vertex $v$ constrain $$\sum_{e \ni v} x_e = \begin{cases} 1 &amp; \text{if } v\in\{s,t\} \\ 0 &amp; \text{otherwise}.\end{cases}$$