We consider the question: "If the zero-framed surgeries on two oriented knots in \(S^3\) are integral homology cobordant, preserving the homology class of the positive meridians, are the knots themselves concordant?" We show that this question has a negative answer in the smooth category, even for topologically slice knots. To show this we first prove that the zero-framed surgery on \(K\) is \(\mathbb{Z}\)-homology cobordant to the zero-framed surgery on many of its winding number one satellites \(P(K)\). Then we prove that in many cases the \(\tau\)- and \(s\)-invariants of \(K\) and \(P(K)\) differ. Consequently neither \(\tau\) nor \(s\) is an invariant of the smooth homology cobordism class of the zero-framed surgery. We also show, that a natural rational version of this question has a negative answer in both the topological and smooth categories, by proving similar results for \(K\) and its \((p,1)\)-cables.