= Solution
Use the local coordinate $t=1/\lambda$. The differential system becomes
$$
\frac{dy}{dt}=-\left(\frac{A_0}{t^4}+\frac{A_1}{t^3}+\frac{A_2}{t^2}\right)y.
$$
For the nonzero quadratic leading term, the coefficient of the local <linear ordinary differential equation> system has a \b[<pole> of order four], so infinity is an <irregular singular point> of <Poincaré rank> \b[three] in this gauge. The <matrix>-valued function $A(\lambda)$ itself has a <pole> of order two; the extra factor $d\lambda/dt=-t^{-2}$ is essential when classifying the equation. If $A_0$ vanishes the degree and these orders decrease. The specified distinct <eigenvalues> $1,-1$ give the nondegenerate unramified case.
The leading <exponential function> factors have actions $q_\pm=\pm\lambda^3/3$, so their difference is $2\lambda^3/3$. Equal exponential magnitudes occur on
$$
\boxed{\operatorname{Re}(\lambda^3)=0,\qquad
\arg\lambda=\frac\pi6+\frac{k\pi}3\quad(k=0,\ldots,5).}
$$
These are commonly called Stokes rays in the sectorial ODE convention. With the action-phase convention used by the wiki's <Stokes line> article they are <anti-Stokes lines>; the same-phase rays instead have
$$
\boxed{\operatorname{Im}(\lambda^3)=0,\qquad\arg\lambda=\frac{k\pi}3.}
$$
Giving both defining equations removes the nomenclature ambiguity; both sets contain six directions.
The standard construction with distinct leading <eigenvalues> uses \b[six <sectorial fundamental solutions> per turn], with basic angular spacing $\pi/3$. They are <fundamental matrices>, each containing two independent column solutions. On the <logarithm> cover choose six consecutive overlapping sectors of width $\pi/3+2\delta$, with $\delta>\varepsilon/2$ small. Their union has angular opening $2\pi+2\delta$, covering the requested $2\pi+\varepsilon$. One often also writes a seventh index to compare the closing sector with the first after a full turn; that is a closing copy, not a seventh independent sector in the six-sector construction. The number is a generic <sectorial fundamental solution> normalization count, not a universal minimum for exceptional systems with vanishing <Stokes matrices>: the exactly diagonal system already has one global elementary <fundamental matrix>.
There is a missing qualification in the requested product identity. <Polynomial> coefficients give an entire invertible <fundamental matrix>, hence trivial actual <monodromy> about a large circle: the circle contracts through <ordinary points>. But the formal <matrix> generally has the form $\widehat G\lambda^\Theta e^Q$, and a positive turn changes its <logarithm> branch by the <formal monodromy> $M_f=e^{2\pi i\Theta}$. This factor cannot be discarded.
To specify the printed product order, let $S_j$ act on column vectors of solution coefficients across the $j$th sector boundary, so adjacent normalized <matrices> satisfy $Y_{j+1}=Y_jS_j^{-1}$. Six transitions give $Y_7=Y_1(S_6\cdots S_1)^{-1}$. The normalized formal branch after the turn is $\widehat Y_1M_f$, so sectorial uniqueness gives $Y_7=Y_1M_f$. Consequently
$$
\boxed{S_6S_5S_4S_3S_2S_1=M_f^{-1}=e^{-2\pi i\Theta}.}
$$
Using <Stokes matrices> for right changes of <fundamental matrices> instead reverses/inverts the corresponding convention; the <formal monodromy> factor remains necessary. The <formal monodromy correction to an entire-system Stokes product> shows that the printed identity holds when \b[$M_f=I$], including the corrected Lax specialization discussed below.
It does not follow from polynomiality alone. For example take
$$
A(\lambda)=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\lambda^2
+\begin{pmatrix}0&1\\-1&0\end{pmatrix}\lambda
+\begin{pmatrix}3/2&1\\2&-3/2\end{pmatrix}.
$$
The coefficient matching in question 2 gives $\Theta=\tfrac12\operatorname{diag}(1,-1)$, hence $M_f=-I$ and the ordered <Stokes matrix> product is $-I$, not $I$, despite having no finite singularity. This is an explicit counterexample to the unqualified final clause.
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