 Solution

There are two conventions for loxodromic Möbius transformations. In the inclusive convention, a transformation conjugate to $z\mapsto \lambda z$ with $|\lambda |\ne 1$ is loxodromic, including a hyperbolic Möbius transformation with $\lambda >0$. Under that convention the assertion about discs is false: $z\mapsto 2z$ preserves the upper half-plane, and a Cayley conjugation gives an invariant ordinary disc.

The question uses the classification in which hyperbolic Möbius transformations and genuinely spiralling loxodromic Möbius transformations are separate. Thus here the multiplier satisfies $|\lambda |\ne 1$ and $\lambda \notin (0,\infty )$. Such a Möbius transformation has two distinct fixed points; after sending them to zero and infinity it acts by the stated dilation and rotation.

For a determinant-one representative, put $\tau =a+d$. Its eigenvalues are $\eta ,{\eta }^{-1}$, with $\eta +{\eta }^{-1}=\tau$, and the normalized multiplier is ${\eta }^2$ or its reciprocal. If $\tau$ is real, the eigenvalues are either real or conjugate unit-modulus numbers. These give the hyperbolic, elliptic, parabolic or identity cases, not a genuine spiral. Conversely, a real eigenvalue or a unit-modulus eigenvalue makes $\tau$ real. Hence the practical criterion in the question's convention isChanging the determinant-one representative by its sign does not affect this criterion.

Finally, conjugate an invariant disc to the upper half-plane. A Möbius transformation preserving that half-plane has a representative in $S{L}_2(\mathbb{R})$: preservation of the boundary gives real coefficients up to a common scalar, and preservation of the chosen side makes the real determinant positive, allowing normalization to one. Its trace is real. Trace is invariant under conjugacy, up to the harmless sign of the representative, so a disc-preserving transformation cannot be genuinely loxodromic. This proves the intended assertion while specifying its necessary terminology.