= Solution
There are two conventions for <loxodromic Möbius transformations>. In the inclusive convention, a transformation conjugate to $z\mapsto\lambda z$ with $|\lambda|\ne1$ is loxodromic, including a <hyperbolic Möbius transformation> with $\lambda>0$. Under that convention the assertion about discs is false: $z\mapsto2z$ 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 $|\lambda|\ne1$ 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 is
$$
\boxed{\operatorname{Im}(a+d)\ne0.}
$$
Changing 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 $SL_2(\mathbb R)$: 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 \b[a disc-preserving transformation cannot be genuinely loxodromic]. This proves the intended assertion while specifying its necessary terminology.
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