$\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\newcommand{\tal}{\tilde\al}\newcommand{\ga}{\gamma}\newcommand{\K}{\mathfrak K}$The answer is yes -- assuming, as done in the comment by the OP, that $x_1=0$ (without this assumption, the answer would clearly be no).

Let
\begin{equation*}
S(\la):=\sup\{L(\la,x,y)\colon (x,y)\in(\R^n)^2,x_1=0\}. \tag{00}\label{00}
\end{equation*}
First, note that for all $\la\in[0,1]$
\begin{equation*}
S(\la)\ge L(\la,0,0)\ge-p, \tag{10}\label{10}
\end{equation*}
where
\begin{equation*}
p:=\min\big\{p_{i,j}e^{-(i-j)}\colon i\in[m],j\in[n]\big\}>0
\end{equation*}
and $[m]:=\{1,\dots,m\}$.

Next,
\begin{equation*}
\sum_{i\in[m]}\al_i x_i + \sum_{j\in[n]}\be_j y_j
\le M_x+M_y,
\end{equation*}
where $M_x:=\max_{1\le i\le m}x_i$ and $M_y:=\max_{1\le j\le n}y_j$, and hence for all $\la\in[0,1]$
\begin{equation*}
L(\la,x,y)\le M_x+M_y-pe^{M_x+M_y}.
\end{equation*}
So, for some real $C>0$, all $\la\in[0,1]$, and all $(x,y)\in(\R^n)^2$ such that $M_x+M_y>C$ we will have $L(\la,x,y)<-p$. So, in view of \eqref{00} and \eqref{10}, for all $\la\in[0,1]$
\begin{equation*}
S(\la)>\sup\{L(\la,x,y)\colon(x,y)\notin E_C\},
\end{equation*}
where
\begin{equation*}
E_C:=\{(x,y)\in(\R^n)^2\colon x_1=0, M_x+M_y\le C\}.
\end{equation*}

Next,
\begin{equation*}
L(\la,x,y)<\sum_{i\in[m]}\al_i x_i+ \sum_{j\in[n]}\be_j y_j.
\end{equation*}
So, using \eqref{10} again, we see that
\begin{equation*}
S(\la)>\sup\{L(\la,x,y)\colon(x,y)\notin E_{p,C}\}, \tag{20}\label{20}
\end{equation*}
where
\begin{equation*}
E_{p,C}:=\Big\{(x,y)\in(\R^n)^2\colon x_1=0, M_x+M_y\le C,\ \\
\sum_{i\in[m]}\al_i x_i+ \sum_{j\in[n]}\be_j y_j\ge-p\Big\}.
\end{equation*}
Letting $\tal_2:=\al_1+\al_2$ and $\tal_i:=\al_i$ for $i=3,\dots,m$, for any $(x,y)\in E_{p,C}$ we have
\begin{equation*}
\begin{aligned}
\al_1 x_2-p&\le\al_1x_2+\sum_{i\in[m]}\al_i x_i+ \sum_{j\in[n]}\be_j y_j \\
& =\al_1x_2+\sum_{i=2}^m\al_i x_i+ \sum_{j\in[n]}\be_j y_j \\
& =\sum_{i=2}^m\tal_i x_i+ \sum_{j\in[n]}\be_j y_j \\
& \le\sum_{i=2}^m\tal_i M_x+ \sum_{j\in[n]}\be_j M_y \\
& =M_x+M_y\le C,
\end{aligned}
\end{equation*}
whence $x_2\le K:=\frac{p+C}{\al_1}$; since $x_2,\dots,x_m$ are interchangeable, similarly we get $x_j\le K$ for all $j=2,\dots,m$; recalling that $x_1=0$, we get
\begin{equation*}
0\le M_x\le K
\end{equation*}
and then
\begin{equation*}
M_y\le C-M_x\le C\le K,
\end{equation*}
for all $(x,y)\in E_{p,C}$.

So, for any $(x,y)\in E_{p,C}$
\begin{equation*}
\begin{aligned}
-p&\le\sum_{i\in[m]}\al_i x_i+ \sum_{j\in[n]}\be_j y_j \\
&=\be_1 y_1+\sum_{i\in[m]}\al_i x_i+ \sum_{j=2}^n\be_j y_j \\
&\le\be_1 y_1+\sum_{i\in[m]}\al_i K+ \sum_{j=2}^n\be_j K
\le \be_1 y_1+2K,
\end{aligned}
\end{equation*}
whence $y_1\ge-\frac{p+2K}{\be_1}$; similarly, $y_j\ge-\frac{p+2K}{\be_j}$ for all $j\in[n]$ and $x_i\ge-\frac{p+2K}{\al_i}$ for all $i\in[m]$; so,
\begin{equation*}
\min_{i\in[m]}x_i\ge-k,\quad
\min_{j\in[n]}y_j\ge-k,
\end{equation*}
where $k:=\frac{p+2K}\ga$ and $\ga:=\min(\min_{i\in[m]}\al_i,\min_{j\in[n]}\be_j)>0$.

So,
\begin{equation*}
E_{p,C}\subseteq\K:=[-k,K]^m\times[-k,K]^n.
\end{equation*}

For each $\la\in[0,1]$, the function $L(\la,\cdot,\cdot)$ is strictly concave and continuous, and thus, in view of \eqref{00} and \eqref{20}, $L(\la,\cdot,\cdot)$ has a unique maximizer $(x_\la,y_\la)$, and $(x_\la,y_\la)$ is in the compact set $\K$, for all $\la\in[0,1]$. $\quad\Box$

proofthat a maximizer exists? $\endgroup$