= Solution
There is a literal inconsistency in the stated <Riemann-Hilbert problem>: a jump <matrix> $S(k)$ tending to zero is incompatible with $\mu_\pm\to I$. Along the real axis the conjugating exponentials are unitary, so the right side of the jump tends to zero while the left side tends to $I$. \b[No normalized solution satisfies the printed decay condition.] The intended condition is $S(k)\to I$, or replacement of $S$ in the jump by $I+S$ when $S$ denotes its decaying part.
For this corrected problem assume a regular, invertible, pole-free solution, differentiable in $x,t$, with the required expansion uniformly at infinity. These conditions matter: arbitrary jump data need not be solvable, and <poles> would contribute extra terms. Write $\sigma=\sigma_3$ and $\mu=I+\mu_1/k+\mu_2/k^2+\cdots$. Matching the nondecaying coefficients in the two differential expressions yields
$$
\boxed{Q=i[\sigma,\mu_1],\qquad A=2i[\sigma,\mu_1]=2Q,\qquad
B=2i[\sigma,\mu_2]-A\mu_1.}
$$
The first expression makes $Q$ off-diagonal.
These matches really establish the differential equations, rather than just formal identities. Let $G=e^{-i(kx+2k^2t)\sigma}S(k)e^{i(kx+2k^2t)\sigma}$. It satisfies $G_x=-ik[\sigma,G]$ and $G_t=-2ik^2[\sigma,G]$. Differentiating $\mu_+=\mu_-G$ consequently shows that
$$
R_x=(\mu_x+ik[\sigma,\mu]-Q\mu)\mu^{-1},\qquad
R_t=(\mu_t+2ik^2[\sigma,\mu]-(kA+B)\mu)\mu^{-1}
$$
have matching boundary values. They therefore extend holomorphically across the contour. The chosen coefficients make both residuals $O(k^{-1})$ at infinity. Their absence of <poles> and <Liouville theorem> imply $R_x=R_t=0$, proving the required <Lax pair> under the corrected hypotheses. This is <matrix NLS reconstruction from a normalized Riemann-Hilbert problem>.
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