= Solution
If $K=K_1\mathbin{\#}K_2$ with both summands nontrivial, the sphere separating the two punctured three-balls in the connected-sum construction is a <splitting sphere of a knot>. If it were trivial, the corresponding one-string tangle would be boundary-parallel and one of $K_1,K_2$ would be the <unknot>. The sphere is therefore nontrivial.
Conversely, a splitting sphere $S$ divides $S^3$ into three-balls $B_1,B_2$, and $K\cap B_i$ is a properly embedded arc. Join its endpoints by a fixed arc on $S$ and push that joining arc slightly into $B_i$; this closes the two tangles to knots $K_1,K_2$. Reversing the construction shows
$$
K=K_1\mathbin{\#}K_2.
$$
If either $K_i$ were unknotted, an innermost-disk argument for a spanning disk of $K_i$ would make the corresponding tangle boundary-parallel, which is exactly the stated triviality condition for $S$. A nontrivial splitting sphere therefore makes both summands nontrivial, so $K$ is composite.
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