This problem, that I will call Subset-Perm-Sum, is NP-complete. Membership is easy: guess the subset non-deterministically and then check.
For hardness one can reduce from 3CNF-SAT in a very similar way to the standard proof of hardness for Subset-Sum.
Let $\varphi$ be an input formula with $v$ variables and $c$ clauses. We will build an instance of Subset-Perm-Sum over $S_{2v+4c}$. For every variable we build $2$ permutations (one that will represent the variable, and one that will represent its negation), and for every clause we will build $2$ permutations as well.
To each clause $C_j, 1\leq j \leq c$ we associate $4$ elements: $2v+4j-k$ for $0 \leq k \leq 3$. To build the $2$ permutations associated to a clause we simply do a cycle on its associated elements. That is, $$ p(C_j) = (2v+4j-3, 2v+4j-2, 2v+4j-1, 2v+4j)$$ (we will add $2$ instance of $p(C_j)$ to the set we want to find a subset-sum later.)
Consider the $i$-th variable, $x_i$, and associate with it the elements $2i-1$ and $2i$ of $S_{2v+4c}$.
To build the permutation $p(x_i)$ of variable $x_i$ simply swap its associated elements ($2i-1$ and $2i$), and multiply by the permutation of each clause it is contained in. Then, in cycle notation you can write:
$$
p(x_i) = (2i-1, 2i) \prod_{j | x_i \in C_j}p(C_j)
$$
Consider now the multiset $M$ that is the union of the $p(x_i)$ and $p(\bar{x_i})$ and two times each $p(C_j)$.
We define the target permutation $t$ as:
$$
t = \prod_{i=1}^v (2i-1, 2i) \cdot \prod_{j=1}^c p(C_j)^3
$$
Claim: There is a subset $X$ of $M$ that composes to the target permutation $t$ (let me write $p(X)=t$) if and only if $\varphi$ is satisfiable.
Assume such an $X$ exists, then we know $X$ contains exactly one out of $p(x_i)$ and $p(\bar{x}_i)$, as it is the only way for the composition of $X$ to include the cycle $(2i-1, 2i)$. Furthermore, we can see that for each clause $C_j \in \varphi$, at least one if variables satisfies it, as to have $p(C_j)^3$ in $p(X)$ we need to have included a $p(\ell_i)$ such that literal $\ell_i$ is in $C_j$. Note that taking the two instances of $p(C_j)$ in $X$ is not enough. Therefore, there is an assignment of variables that satisfies every clause.
For the backward direction, let $\sigma$ be a satisfying assignment of $\varphi$. If $\sigma(x_i) = 1$ then we let $X$ contain $p(x_i)$ and otherwise we let it contain $p(\bar{x}_i)$. For each clause $C_j$, we know there are between $0 \leq r_j \leq 2$ variables on it that are not satisfied by $\sigma$, and we add $r_j$ occurrences of $p(C_j)$ to $X$, so this way $p(X)$ includes $p(C_j)^3$. It is clear that the composition of $X$ equals $t$.
