$\newcommand{\si}{\sigma}\newcommand{\al}{\alpha}\newcommand\la\lambda$We have
\begin{equation*}
|\al|=h^{-1-d/2}|g(1)-g(0)|\le h^{-1-d/2}\sup_{s\in[0,1]}|g'(s)|,
\end{equation*}
where
\begin{equation*}
g(s):=\frac{G((\sqrt h\si(s))^{-1}y(s))}{\sqrt{\det a(s)}},
\end{equation*}
$y(s):=u-hb(s)$, $u:=x'-x$,
$\si(s):=(1-s)\si(t,x)+s\si(t,x')$, $b(s):=(1-s)b(t,x)+sb(t,x')$, $a(s):=(1-s)a(t,x)+sa(t,x')$.
For $s\in[0,1]$,
\begin{equation*}
\frac{g'(s)}{g(s)}=g_1+g_2+g_3,
\end{equation*}
where
\begin{equation*}
g_1:=-\frac{\det a(1)-\det a(0)}{2a(s)}, \quad
g_2:=-(b(1)-b(0))^\top a(s)^{-1}y(s),
\end{equation*}
\begin{equation*}
g_3:=\frac1h\, y(s)^\top a(s)^{-1}(a(1)-a(0))a(s)^{-1}y(s).
\end{equation*}
We have $\la^{-1}I\le a(s)\le\la I$ and hence
\begin{equation*}
\la^{-1}\le |a(s)|\le\la, \quad \det a(s)\ge\la^{-d},
\end{equation*}
and, in view of this answer, $|\det a(1)-\det a(0)|\le\la^{d-1}d\,|a(1)-a(0)|\le\la^{d-1}d\,L|u|^\eta$. So,
\begin{equation*}
|g_1|\ll|u|^\eta;
\end{equation*}
here and in what follows, $A\ll B$ and $B\gg A$ mean that $A\le CB$ for some real $C>0$ depending only on $d,\la,L,T$ -- given that $h\in(0,T]$. Next,
\begin{equation*}
|g_2|\le2L\la(|u|+hL)\ll|u|+h,
\end{equation*}
\begin{equation*}
|g_3|\le\frac1h\,\la^2 L|u|^\eta (|u|+hL)^2\ll\frac{|u|^{\eta+2}}h+|u|^\eta,
\end{equation*}
since $h\in(0,T]$.
So,
\begin{equation*}
\frac{h^{1+d/2}|\al|}{g(s)}\ll |u|^\eta+|u|+h+\frac{|u|^{\eta+2}}h.
\end{equation*}
Next,
\begin{equation*}
g(s)\le\la^{d/2}\exp\Big(-\frac{|u|^2/2-(Lh)^2}{2\la h}\Big)
\ll\exp\Big(-\frac{|u|^2}{4\la h}\Big)
\ll h^{d/2}e^{-c|u|^2/h}p_c(h,x,x')
\end{equation*}
for $c:=\frac1{6\la}$. So,
\begin{equation*}
\frac{|\al|}{h^{-1+\eta/2} p_c (h,x,x')}
\ll \Big(|u|^\eta+|u|+h+\frac{|u|^{\eta+2}}h\Big)h^{-\eta/2}e^{-c|u|^2/h}. \tag{10}\label{10}
\end{equation*}
Further,
\begin{equation*}
|u|^\eta h^{-\eta/2}e^{-c|u|^2/h}=(|u|^2/h)^{\eta/2}e^{-c|u|^2/h}
\le\max_{w\ge0}w^{\eta/2}e^{-cw}\ll1; \tag{20}\label{20}
\end{equation*}
\begin{equation*}
|u| h^{-\eta/2}e^{-c|u|^2/h}=h^{(1-\eta)/2}(|u|^2/h)^{1/2}e^{-c|u|^2/h}
\le T^{(1-\eta)/2}\max_{w\ge0}w^{1/2}e^{-cw}\ll1 \tag{30}\label{30}
\end{equation*}
provided that
\begin{equation*}
\eta\le1; \tag{$*$}\label{*}
\end{equation*}
\begin{equation*}
h\,h^{-\eta/2}e^{-c|u|^2/h}\le h^{1-\eta/2}\ll1,
\end{equation*}
again provided \eqref{*};
\begin{equation*}
\frac{|u|^{\eta+2}}h h^{-\eta/2}e^{-c|u|^2/h}=(|u|^2/h)^{\eta/2+1}e^{-c|u|^2/h}
\le\max_{w\ge0}w^{\eta/2+1}e^{-cw}\ll1. \tag{40}\label{40}
\end{equation*}
Now the desired inequality
\begin{equation*}
|\al|\ll h^{-1+\eta/2} p_c (h,x,x') \tag{50}\label{50}
\end{equation*}
follows from \eqref{10}, \eqref{20}, \eqref{30}, and \eqref{40}, provided \eqref{*}. $\quad\Box$
Without condition \eqref{*}, inequality \eqref{50} will fail to hold in this generality.
E.g., suppose that for all $t,x$ we have $\si(t,x)=I$ and $|b(t,x)|\in[0,1]$, so that $a(t,x)=I$ and your conditions 1 and 2 hold with $\la=1$, $L=2$, and any real $\eta>0$. Suppose also that $x=0$, $|x'|=\sqrt h$, $b(t,0)=0$, and $b(t,x')=-x'/|x'|$. Then
\begin{equation}
\frac{|\al|}{h^{-1+\eta/2} p_c (h,x,x')}\asymp h^{1/2-\eta/2}\to\infty
\end{equation}
as $h\downarrow0$ if $\eta>1$. $\quad\Box$
Best Answer
For real $u,v$, let \begin{equation*} Q(u,v):=P(Y_1>u\sqrt2,Y_2>v\sqrt2)=\int_u^\infty dz\,\varphi(z)\Big(1-\Phi\Big(\frac{2v-z}{\sqrt3}\Big)\Big), \end{equation*} where $\varphi$ and $\Phi$ are the standard normal pdf and cdf, respectively.
The question in the OP can be restated as follows: show that \begin{equation*} r(u,v):=r_t(u,v):=\frac{Q(u-t,v-t)}{Q(u,v)} \tag{10}\label{10} \end{equation*} is increasing in real $u$ and in real $v$ for each $t\in(0,1/\sqrt2)$. Since $u$ and $v$ are interchangeable, it is enough to shown that \begin{equation*} r(v):=r(u,v) \end{equation*} is increasing in real $v$ for each $t\in(0,1)$.
This will be done by using so-called l'Hospital-type rules for monotonicity. Indeed, consider the "derivative ratio" for the ratio in \eqref{10} with the extra constant factor $e^{t^2/2}$: \begin{equation*} r_1(v):=e^{t^2/2}\frac{Q_v(u-t,v-t)}{Q_v(u,v)}, \end{equation*} where the subscript $_v$ denotes the partial derivative in $v$. We have \begin{equation*} Q_v(u,v)=-\int_u^\infty dz\,\varphi(z)\,\frac2{\sqrt3}\,\varphi\Big(\frac{2v-z}{\sqrt3}\Big) =-\frac{\varphi(v)}2\, \left(\text{erf}\left(\frac{v-2 u}{\sqrt{6}}\right)+1\right). \end{equation*} So, \begin{equation*} r_1(v)=\frac{f_1(v)}{g_1(v)}, \tag{20}\label{20} \end{equation*} where \begin{equation*} f_1(v):=e^{t v} \left(\text{erf}\left(\frac{t-2 u+v}{\sqrt{6}}\right)+1\right),\quad g_1(v):=\text{erf}\left(\frac{v-2 u}{\sqrt{6}}\right)+1. \end{equation*} Next, consider the "derivative ratio" for the ratio in \eqref{20}: \begin{equation*} r_2(v):=\frac{f'_1(v)}{g'_1(v)}=\frac{f_2(v)}{g_2(v)}, \tag{30}\label{30} \end{equation*} where \begin{equation*} f_2(v):=\sqrt{6 \pi } t \left(\text{erf}\left(\frac{t-2 u+v}{\sqrt{6}}\right)+1\right)+2 e^{-(t-2 u+v)^2/6},\quad g_2(v):=2 e^{-t v-(v-2 u)^2/6}. \end{equation*} Further, consider the third (and final) "derivative ratio" for the ratio in \eqref{30}: \begin{equation*} r_3(v):=\frac{f'_2(v)}{g'_2(v)}=\frac{e^{-t (t-4 (u+v))/6}\, (-2 t-2 u+v)}{3 t-2 u+v}. \tag{40}\label{40} \end{equation*} Note that for all real $z$ \begin{equation} r'_3(2u+z)=\frac{t \left(2 z^2+2 t z+15-12 t^2\right) e^{-t (t-4 (3 u+z))/6}}{3 (3 t+z)^2}, \end{equation} which is obviously $>0$ for all $t\in(0,1)$ if $z\ne-3t$.
So, $r_3$ is increasing on $(-\infty,2u-3t)$ and on $(2u-3t,\infty)$. Also, $g_2>0$. Also, $g'_2>0$ on $(-\infty,2u-3t)$ and $g'_2<0$ on $(2u-3t,\infty)$.
So, by lines 1 and 3 of Table 1.1, (the continuous function) $r_2$ is down-up on $(-\infty,2u-3t]$ and up-down on $[2u-3t,\infty)$ -- that is, (i) there is some $a\in[-\infty,2u-3t]$ such that $r_2$ is decreasing on $(-\infty,a]$ and increasing on $[a,2u-3t]$ and (ii) there is some $b\in[2u-3t,\infty]$ such that $r_2$ is increasing on $[2u-3t,b]$ and decreasing on $[b,\infty)$.
So, $r_2$ is down-up-down on $(-\infty,\infty)$. But $r_2$ is a positive function such that $r_2(v)\to0$ as $v\to-\infty$ and $r_2(v)\to\infty$ as $v\to\infty$. So, $r_2$ is increasing on $(-\infty,\infty)$.
But $f_1(v)\to0$ and $g_1(v)\to0$ as $v\to-\infty$. So, by Proposition 4.1 of the cited paper, $r_1$ is increasing on $(-\infty,\infty)$.
Also, clearly $Q(u,v)\to0$ as $v\to\infty$. So, again by Proposition 4.1 of the cited paper, $r(u,v)$ is increasing in $v\in (-\infty,\infty)$. $\quad\Box$