= Solution
Work locally in $z$ where $u\ne0$, and take the <matrix> entries literally as they appear in the original PDF. At fixed $z$, abbreviate
$$
D=\operatorname{diag}(1,-1),\quad U=u,\quad P=u',\quad
V=-\frac{zw}{u},\quad R=\frac{2u'w}{u^2},\quad d=w+\frac z2.
$$
Then $A=D\lambda^2+\left(\begin{smallmatrix}0&U\\V&0\end{smallmatrix}\right)\lambda+\left(\begin{smallmatrix}d&P\\R&-d\end{smallmatrix}\right)$. Define
$$
h=d+\frac{UV}{2},\qquad \theta=\frac{UR+VP}{2},\qquad
\rho=\frac{UV(d+h)}4-\frac{PR}{2}.
$$
The literal coefficients give
$$
\boxed{h=\frac z2+\left(1-\frac z2\right)w,\qquad
\theta=\frac{(2-z)u'w}{2u}.}
$$
In general neither $h=z/2$ nor $\theta=0$ is valid for the printed system.
The normalized formal <fundamental matrix> is
$$
\boxed{\widehat Y=\left(I+\frac{G_1}{\lambda}+\frac{G_2}{\lambda^2}+\cdots\right)
\lambda^{\theta D}\exp\left[\left(\frac{\lambda^3}{3}+h\lambda\right)D\right],}
$$
with a chosen <logarithm> branch. In particular the diagonal normal form has $\Lambda_1=0$, $\Lambda_2=hD$, $\Lambda_3=\theta D$. Its first two prefactor coefficients are
$$
\boxed{G_1=\begin{pmatrix}\rho&-U/2\\V/2&-\rho\end{pmatrix},}
$$
$$
\boxed{G_2=\begin{pmatrix}
\rho^2/2-UV/8+h\theta/2&(U\rho-P)/2\\
(V\rho+R)/2&\rho^2/2-UV/8-h\theta/2
\end{pmatrix}.}
$$
These formulas include the diagonal terms required by the <derivative> of the prefactor. Setting every diagonal prefactor coefficient to zero while retaining only the displayed exponent would generally fail the <differential equation>.
Here is the coefficient calculation and the full recursion. Write $A_1=\left(\begin{smallmatrix}0&U\\V&0\end{smallmatrix}\right)$, $A_2=\left(\begin{smallmatrix}d&P\\R&-d\end{smallmatrix}\right)$ and $G_0=I$, with negative-index $G_j=0$. Substitution into the <gauge transformation of a linear differential system> gives, for $m\ge1$,
$$
[D,G_m]=-A_1G_{m-1}-A_2G_{m-2}+G_{m-2}hD
+G_{m-3}[\theta D-(m-3)I].
$$
At $m=1$, the <commutator> determines $(G_1)_{12}=-U/2$, $(G_1)_{21}=V/2$ and the absence of a linear diagonal term. The diagonal part at $m=2$ gives $h=d+UV/2$. The next diagonal part gives $\theta=(UR+VP)/2$. At $m=4$ diagonal solvability gives $-\rho=PR/2-UV(d+h)/4$; the next order fixes the two diagonal entries of $G_2$ above. The off-diagonal entries are fixed at each stage because $[D,X]_{12}=2X_{12}$ and $[D,X]_{21}=-2X_{21}$. Later diagonal solvability fixes the remaining normalization coefficients. Thus this is a complete formal solution, not a convergence assertion. It is the <formal cubic-exponential two-by-two system> expansion. The printed hint's second prefactor term must have denominator $\lambda^2$, and the remainder after the $1/\lambda$ term of the diagonal expansion must start at $1/\lambda^2$.
Now impose an <isomonodromic deformation> in the full irregular sense: its <Stokes matrices> and formal exponent are fixed in $z$, with the leading formal normalization fixed. Sectorial <matrices> satisfy $Y_{j+1}=Y_jC_j$ with $(C_j)_z=0$, so $B_j=(Y_j)_zY_j^{-1}$ is the same on overlaps. Since the spectral system has no finite singularities, its <fundamental matrix> and inverse are entire in $\lambda$, making $B$ entire. Differentiating the formal expression, using $\theta_z=0$, gives
$$
B=G_zG^{-1}+G(h_z\lambda D)G^{-1}
=h_z\lambda D+h_z[G_1,D]+O(\lambda^{-1}).
$$
But $[G_1,D]=A_1$. The sectorial cover makes this a global growth bound, so <Liouville's theorem> applied to $B-h_z(\lambda D+A_1)$ gives the literal consequence
$$
\boxed{B=h_z\begin{pmatrix}\lambda&u\\-zw/u&-\lambda\end{pmatrix}.}
$$
An arbitrary parameter dependence of $u,w$ need not be isomonodromic; this analytic argument presupposes the stated constancy of generalized <monodromy> data.
The requested <matrix> would require $h_z=1/2$ from its diagonal and upper entry, and then its lower entry would require $(z-2)w=0$. On an open deformation domain this forces $w\equiv0$. Thus that requested expression is not the general deformation <matrix> for the printed coefficients.
There is a consistent nearby correction: replace the lower entry of $A_1$ by $-2w/u$. Then $V=-2w/u$, so $h=z/2$ and $\theta=0$, and the same derivation proves exactly
$$
\boxed{Y_z=\begin{pmatrix}\lambda/2&u/2\\-w/u&-\lambda/2\end{pmatrix}Y.}
$$
For this corrected system use the same prefactor formulas with $V=-2w/u$: $\rho=-w(w+z)/2-w(u'/u)^2$ and both diagonal entries of $G_2$ are $\rho^2/2+w/4$. Compatibility determines the allowed $u,w$ in the next solution. This correction is stated explicitly rather than silently substituting it for the original PDF.
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