It might help you to look at each side separately.
Starting with the left-hand side
lhs = Sum[(p/(1 - p))^s*(q/(1 - q))^s*
Binomial[n,
s]*(Binomial[m - 1, s]*(p*q*(m + n) + (2*m - 1)*(-p - q + 1))), {s, 0, n}]
-Hypergeometric2F1[1 - m, -n, 1,
(pq)/((-1 + p)(-1 + q))] + 2*mHypergeometric2F1[1 - m, -n,
1, (pq)/((-1 + p)(-1 + q))] + pHypergeometric2F1[1 - m, -n, 1,
(pq)/((-1 + p)(-1 + q))] - 2*mpHypergeometric2F1[1 - m,
-n, 1, (pq)/((-1 + p)
(-1 + q))] + qHypergeometric2F1[1 - m, -n, 1,
(pq)/((-1 + p)*(-1 + q))] - 2*mqHypergeometric2F1[1 - m,
-n, 1, (pq)/((-1 + p)
(-1 + q))] + mpqHypergeometric2F1[1 - m,
-n, 1, (pq)/((-1 + p)*
(-1 + q))] + npqHypergeometric2F1[1 - m,
-n, 1, (pq)/((-1 + p)*
(-1 + q))]
lhs = lhs // Simplify
(-1 + p + m*(2 + p*(-2 + q) -
2*q) + q + npq)* Hypergeometric2F1[1 - m, -n, 1,
(pq)/((-1 + p)(-1 + q))]
And for the right-hand side
rhs = Sum[(p/(1 - p))^s*(q/(1 - q))^s*
Binomial[n,
s]*((-(-p - q + 1))*Binomial[m - 2, s] + m*p*q*Binomial[m, s] +
m*(-p - q + 1)*(Binomial[m - 2, s] + Binomial[m, s])), {s, 0, n}]
-Hypergeometric2F1[2 - m, -n, 1,
(pq)/((-1 + p)(-1 + q))] + mHypergeometric2F1[2 - m, -n, 1,
(pq)/((-1 + p)(-1 + q))] + pHypergeometric2F1[2 - m, -n, 1,
(pq)/((-1 + p)(-1 + q))] - mpHypergeometric2F1[2 - m, -n,
1, (pq)/((-1 + p)(-1 + q))] + qHypergeometric2F1[2 - m, -n, 1,
(pq)/((-1 + p)(-1 + q))] - mqHypergeometric2F1[2 - m, -n,
1, (pq)/((-1 + p)(-1 + q))] + mHypergeometric2F1[-m, -n, 1,
(pq)/((-1 + p)(-1 + q))] - mpHypergeometric2F1[-m, -n, 1,
(pq)/((-1 + p)(-1 + q))] - mqHypergeometric2F1[-m, -n, 1,
(pq)/((-1 + p)(-1 + q))] + mpqHypergeometric2F1[-m, -n,
1, (pq)/((-1 + p)*(-1 + q))]
rhs = rhs // Simplify
(-(-1 + m))(-1 + p + q)
Hypergeometric2F1[2 - m, -n, 1,
(pq)/((-1 + p)(-1 + q))] + m*(-1 + p)(-1 + q)
Hypergeometric2F1[-m, -n, 1,
(pq)/((-1 + p)(-1 + q))]
rhs = rhs // FullSimplify
(-1 + p + m*(2 + p*(-2 + q) -
2*q) + q + npq)* Hypergeometric2F1[1 - m, -n, 1,
(pq)/((-1 + p)(-1 + q))]
The simplified forms are the same
lhs === rhs
True