Let $$a(n,m)=\sum\limits_{i=0}^{m-1}\binom{n+i-1}{i}\beta^i$$ $$b(n,m,k)=\left(\sum\limits_{i=0}^{m-1}\binom{n+i-1}{i}\binom{m-i+k}{k}\beta^i\right)+\sum\limits_{j=1}^{k+1}\binom{n+m-2}{m-1}\beta^{m+j-1}$$ $$c(n,m)=a(n,m)\alpha^n+\sum\limits_{k=1}^{n}b(n-k+1,m,k)\alpha^{n-k}$$ I conjecture that for $n\geqslant0$, $m>0$ $$c(n,m)=\binom{n+m}{m}f(n,m)$$ Here is the PARI prog to verify this conjecture:
f(n, m) = if(n==0 || m==0, (n==0)*(1 - a^(n+1))/(1-a)+(m==0)*(1 - b^(n+1))/(1-b), a*n/(n+m)*f(n-1, m)+b*m/(n+m)*f(n, m-1)+1)
a1(n, m) = sum(i=0, m-1, binomial(n+i-1, i)*b^i)
b1(n, m, k) = sum(i=0, m-1, binomial(n+i-1, i)*binomial(m-i+k, k)*b^i) + sum(j=1, k+1, binomial(n+m-2, m-1)*b^(m+j-1))
c(n, m) = a1(n, m)*a^n + sum(k=1, n, b1(n-k+1, m, k)*a^(n-k))
test(n, m) = c(n, m)==binomial(n+m,m)*f(n,m)
But is there a way to prove it?