This answer was deleted and then updated.
Let $$g(n,m,k)=\sum\limits_{i=0}^{m}\binom{n+i}{i}\binom{m-i+k}{k}\beta^i$$ $$h(n,m)=\sum\limits_{k=0}^{n}b(k,m,n-k)\alpha^k$$ I conjecture that $$h(n,m)=\binom{n+m}{m}f(n,m)$$ Here is the PARI prog to verify this conjecture:
f(n, m) = if(n==0 || m==0, (m==0)*(1 - a^(n+1))/(1-a)+(n==0)*(1 - b^(m+1))/(1-b)-(n==0 && m==0), a*n/(n+m)*f(n-1, m)+b*m/(n+m)*f(n, m-1)+1)
g(n, m, k) = sum(i=0, m, binomial(n+i, i)*binomial(m-i+k, k)*b^i)
h(n, m) = sum(k=0, n, g(k, m, n-k)*a^k)
test(n, m) = h(n, m)==f(n, m)*binomial(n+m, m)
n=6; x=sum(i=0, n, sum(j=0, n, test(i, j)))
But is there a way to prove it?