Math Overflow seems to have a dearth of low dimensional topology, but this seems like an interesting question. Let $L$ be a $2$-component link in $S^3$. Suppose that there is a framed arc joining the two components such that surgery along that arc yields an unknot. (In other words, fatten the arc to a long thin rectangle where two of its edges are subarcs of the link components. Delete these arcs and replace them with the other two edges of the rectangle.) I wonder how restrictive this condition is. For example, can one prove that there are links which don't have this property? I don't really have a good feel for what is possible. 


One way to construct an example is to start with an unknotted surface with one boundary component. Now attach a band to the boundary of this surface that travels through all of the surfaces holes. This is now a 2-component link. Snipping the band can be realized by surgery along a transverse arc, and yields the boundary of the original surface, which is an unknot.



Surgery along an arc connecting the components of a $2$-component link gives the unknot
One source of restrictions is the Montesinos trick: if you take the branched double cover of $L$, then a small neighborhood of the framed arc lifts to a solid torus because it intersects $L$ in two small arcs. Replacing those small arcs with the ones coming from the surgery produces another solid torus in the new branched cover, so the two branched covers are related by a Dehn surgery. If the surgery produces the unknot, then this means that the branched double cover $\Sigma(L)$ has to be Dehn surgery on a knot in $S^3$.


For example, the above implies that $H\_1(\Sigma(L))$ must be cyclic. If $K$ is a knot with determinant $d > 1$, then $H\_1(\Sigma(K))$ has order $d$. The connected sum $L = K \# T(2,2d)$ is a 2-component link with branched double cover $\Sigma(K) \# L(2d,1)$, where $T(2,2d)$ is the $(2,2d)$-torus link and $L(2d,1)$ a lens space. Its first homology is $H\_1(\Sigma(K)) \oplus \mathbb{Z}/(2d)$, and since this is not cyclic, no surgery on $L$ will produce the unknot.


This also reproves the result of Scharlemann which Ian Agol mentioned, because if L is a split link $L\_1 \sqcup L\_2$ then its branched cover is $\Sigma(L\_1) \# \Sigma(L\_2) \# S^1\times S^2$. Then by Gabai's proof of the Poenaru conjecture (which came after Scharlemann's paper), we must have $\Sigma(L\_1)=\Sigma(L\_2)=S^3$, and so $L\_1$ and $L\_2$ are both unknots.


