It’s convenient to define $\alpha_i=0$ and $\beta_i=0$ when $i$ is out of bounds.
Let $a_i$ be the coefficient of $x^i$. The only solutions to $k+j-2s=0$ with $s\le\min(j,k)$ are $k=j=s$, so $a_0=\sum_{i\ge 0}\alpha_i\beta_ic_i(i,i)\;.$
If $k+j-2s=1$, either $k=j+1$ and $s=j$, or $j=k+1$ and $s=k$, so
$a_1=\sum_{i\ge 0}\bigg(\alpha_i\beta_{i+1}c_i(i+1,i)+\alpha_{i+1}\beta_ic_i(i,i+1)\bigg)\;.$
More generally, if $k+j-2s=d$, the only possibilities are
$\begin{align*}k&=s,j=s+d\;;\\ k&=s+1,j=s+d-1\;;\\ k&=s+2,j=s+d-2\;;\\ &\;\vdots\\ k&=s+d,j=s\;, \end{align*}$
and
$a_d=\sum_{s\ge 0}\sum_{i=0}^d\alpha_{s+i}\beta_{s+d-i}c_s(s+d-i,s+i)$
As a quick check, this yields
$a_{m+n}=\sum_{s\ge 0}\sum_{i=0}^{m+n}\alpha_{s+i}\beta_{s+m+n-i}c_s(s+m+n-i,s+i)\;,$
and it’s not hard to see that the only non-zero term is $\alpha_m\beta_nc_0(n,m)$, when $s=0,i=m$, as it should be: the indices of $\alpha$ and $\beta$ sum to $s+m+n$, so if $s>0$, at least one of $\alpha_{s+i}$ and $\beta_{s+m+n-i}$ is zero, and the same is true if $s=0$ and $i\ne m$.