= Solution
Subtract a <steady state> to obtain $e(t)=s(t)-s_*$ with $\dot e=Ae$, hence $e(t)=e^{At}e(0)$. The <steady state> attracts every initial condition exactly when
$$
\boxed{\operatorname{Re}\mu<0\quad\text{for every eigenvalue }\mu\text{ of }A.}
$$
In other words, $A$ must be a <Hurwitz stable matrix>. To see sufficiency without assuming diagonalizability, each block in its <Jordan normal form> contributes terms of the form $t^k e^{\mu t}$ with finite $k$. Negative real part makes every such term tend to zero, so the whole <matrix exponential> tends to zero.
For necessity, an eigenmode with positive real part grows, and a mode with zero real part persists or oscillates instead of decaying. For a complex <eigenvalue> of a real matrix, take the real and imaginary parts of its eigenvector to obtain real nondecaying solutions. Thus any <eigenvalue> in the closed right half-plane prevents convergence of all deviations. Physical <density operators> span the trace-one affine space, so convergence from all physical initial states also forces decay on every traceless direction.
This gives global <asymptotic stability> for the affine linear system and uniqueness of its <steady state>. A merely negative-semidefinite symmetric part proves only nonincrease of a particular norm, not this necessary-and-sufficient condition.
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