= Solution
For an invertible <fundamental matrix>, differentiating $Y_\lambda=AY$ in $z$ and $Y_z=BY$ in $\lambda$ gives $Y_{\lambda z}=(A_z+AB)Y$ and $Y_{z\lambda}=(B_\lambda+BA)Y$. Thus the necessary and sufficient compatibility identity is
$$
\boxed{F:=A_z-B_\lambda+[A,B]=0.}
$$
Here both <derivatives> must use the actual deformation variable $z$; the printed $t$ has not been defined. The <isomonodromic deformation> identity can be checked as a <polynomial> in $\lambda$, without assuming the conclusion.
First retain the coefficients actually printed in the PDF. Write $p=u'$, $d=w+z/2$, and introduce the two residuals
$$
K=w'-\frac{2u'w}{u},\qquad H=u''+du.
$$
Direct <matrix multiplication> and differentiation give
$$
\begin{aligned}
F={}&\lambda^2\begin{pmatrix}0&0\\(2-z)w/u&0\end{pmatrix}
+\lambda\begin{pmatrix}(z-2)w/2&0\\
-(w+zw')/u+(z+2)u'w/u^2&-(z-2)w/2\end{pmatrix}\\
&+\begin{pmatrix}K&H\\2wH/u^2+2u'K/u^2&-K\end{pmatrix}.
\end{aligned}
$$
For example, the quadratic lower entry is the difference between $[D,B_0]_{21}=2w/u$ and the contribution $(A_1)_{21}=-zw/u$; it does not cancel. The constant lower entry follows by differentiating $2u'w/u^2$ and collecting it as $2wH/u^2+2u'K/u^2$. These entries also provide independent checks of the two most important cancellations.
Work on an open connected deformation domain where $u\ne0$, as the <matrices> require. The quadratic entry forces $(2-z)w=0$ throughout the domain, hence $w\equiv0$ by continuity, including a possible point $z=2$. All remaining entries then vanish exactly when $u''+(z/2)u=0$. Thus the literal answer is
$$
\boxed{w\equiv0,\qquad u''+\frac z2u=0.}
$$
This is a rescaled <Airy function> equation: with $\xi=-2^{-1/3}z$, the equation is $u_{\xi\xi}-\xi u=0$, so $u=c_1\operatorname{Ai}(\xi)+c_2\operatorname{Bi}(\xi)$ on a domain avoiding its zeros. In particular, choose the local solution with $u(1)=u'(1)=1$ and $w=0$. It gives compatible <matrices> but violates the printed $w=u'$. This proves that the requested assertion is false for the original coefficients, independently of the undefined $t/z$ in the printed <differential equation>.
For completeness, use the explicitly corrected lower coefficient $(A_1)_{21}=-2w/u$ from the preceding solution. The same multiplication gives the much simpler residual
$$
F=\lambda\begin{pmatrix}0&0\\-2K/u&0\end{pmatrix}
+\begin{pmatrix}K&H\\2wH/u^2+2u'K/u^2&-K\end{pmatrix}.
$$
Consequently compatibility is equivalent to $K=H=0$. The first equation integrates without dividing by $w$:
$$
\left(\frac{w}{u^2}\right)'=\frac1{u^2}\left(w'-\frac{2u'w}{u}\right)=0.
$$
The <constant of integration> $C$ therefore satisfies $w=Cu^2$, and the second equation becomes
$$
\boxed{w=Cu^2,\qquad u''+Cu^3+\frac z2u=0.}
$$
Conversely these two equations give $K=H=0$, and substitution makes every residual entry vanish, proving sufficiency as well as necessity. This is the <quadratic spectral Lax pair for zero-parameter Painlevé II>. For $C\ne0$, choose constants $b,a$ with $b^3=-2$ and $Ca^2b^2=-2$, set $z=bx$, $u=aq(x)$, and obtain $q''=2q^3+xq$, the zero-parameter <Painlevé II equation>. For $C=0$ it reduces to the <Airy function> equation. Even with this correction, the conclusions are $w=Cu^2$ and the coefficient $z/2$, rather than the incompatible expressions printed in the question.
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