Do actions of BS(1,n) on finite sets factor through abelian quotients?
The answer is no. Since every finite group embeds in $S_m$ for some $m$, this is equivalent to asking if every finite quotient of $BS(1,n)$ is abelian. The finite quotients of $BS(1,n)$ are all the quotients of groups of the form $\mathbb{Z}/s\mathbb{Z} \rtimes \mathbb{Z}/r\mathbb{Z}$ where the action is multiplication by $n$ and $n^r \equiv 1 \bmod s$. See this paper by Matei and Suciu for example.
No, these groups have many finite non-abelian quotients. Recall that the Baumslag-Solitar group $B(1,n)$ has a presentation $$ B(1,n)=\langle t,a\,|\,tat^{-1}=a^n\rangle. $$ A homomorphism $\Phi: B(1,n)\to S_m$ amounts to a pair $(\tau,\alpha)$ of permutations, where $\Phi(a)=\alpha$ and $\Phi(t)=\tau$, such that $\tau$ conjugates $\alpha$ to $\alpha^n$.Observe that in the symmetric group $S_m$, a cycle $\alpha$ of length $k$ is conjugate to its $n$th power $\alpha^n$ provided $n$ and $k$ are relatively prime, and $\alpha^n\ne \alpha$ for $k$ not dividing $n-1$, so that the conjugating permutation $\tau$ does not commute with $\alpha$. This yields examples of homomorphisms with non-abelian image.
For a specific example, take $n=2, k=3$ and let $\Phi: B(1,2)\to S_3$ map the
generators $t$ and $a$ of
$$
B(1,2)=\langle t,a\,|\,tat^{-1}=a^2\rangle
$$
into permutations $\tau=\Phi(t)=(2,3)$ and $\alpha=\Phi(a)=(1,2,3)$ in the cycle notation. Since $\Phi(t)=(2,3)$ conjugates $\Phi(a)=(1,2,3)$ into its square $\Phi(a)^2=(1,3,2)$ and moreover, $(2,3)$ and $(1,2,3)$ generate $S_3$, this defines a group homomorphism whose image is $S_3$.
Another way to see the answer is no: the dihedral group $D_{2k}$ (sometimes written $D_k$) of order $2k$ is generated by $a$ ('rotation through $2\pi/k$') and $t$ (a reflection) and satisfies the relation $tat^{-1}=a^{-1}=a^{k-1}$. This is clearly a finite non-abelian group provided $k>2$, and taking $k=n+1$, we see that $D_{2k}$ is an image of $\mathrm{BS}(1,n)$ under the obvious map.
Finally, since $D_{2k}$ acts on the vertices of a regular $k$-gon, $D_{2k}$ embeds in $S_m$ for any $m\geq k$. This gives a negative answer to the question provided $n\geq 2$ and $m\geq 3$.