= Solution
Let $P\mathbf x=(\mathbf n\cdot\mathbf x)\mathbf n$ be the axial <orthogonal projection> and $N\mathbf x=\mathbf n\times\mathbf x$. Then $NP=PN=0$, while the <vector> triple-product identity gives $N^2=P-I$. The map is $aP+a(I-P)+bN$. Its inverse is $P/a$ on the axis. On the perpendicular <plane>,
$$
(aI-bN)(aI+bN)=a^2I-b^2N^2=(a^2+b^2)I.
$$
Thus the <inverse of an axis-angle linear map> is
$$
\mathbf x=\frac{P\mathbf x'}a+\frac{a(I-P)\mathbf x'-bN\mathbf x'}{a^2+b^2},
$$
or, in the requested <vector> form,
$$
\boxed{\mathbf x=\frac{a\mathbf x'-b(\mathbf n\times\mathbf x')+(b^2/a)(\mathbf n\cdot\mathbf x')\mathbf n}{a^2+b^2}.}
$$
The denominators are nonzero because $a,b>0$. Multiplication on each invariant component verifies the <matrix inverse> without choosing a coordinate frame.
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