= Solution
For $k=a+ib$, $\operatorname{Re}\omega=a^2-b^2+\beta b$. The upper contour is therefore the graph
$$
\boxed{L:\quad b=\frac{\beta+\sqrt{\beta^2+4a^2}}2,\quad -\infty<a<\infty,}
$$
oriented from left to right. Between the real axis and $L$, $b\geq0$ and $\operatorname{Re}\omega\geq0$. Hence $e^{ikx-\omega(t-s)}$ is bounded by $e^{-bx}$ for $0\leq s\leq t$. The boundary transforms are entire in $k$; with the standard smooth, compatible boundary data, <integration by parts> in the time integral gives the required polynomial bounds. Closing the intervening sectors and using exponential decay for $x>0$ deforms their real-line integrals onto $L$. Small endpoint truncations, followed by dominated limits, handle the sector boundaries.
For $k'= -k+i\beta$, $\omega(k')=\omega(k)$ and $\operatorname{Im}k'\leq0$ on $L$. Thus the already valid transformed equation at $k'$ says
$$
\widehat u(k',t)=e^{-\omega t}\widehat u(k',0)+ikf_0-f_1.
$$
It eliminates the unknown boundary derivative through
$$
\boxed{f_1=e^{-\omega t}\widehat u(k',0)+ikf_0-\widehat u(k',t).}
$$
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