Let $a_1$ and $a_2$ be real numbers. Let $n_1$ and $n_2$ be positive integers. Finally let $\theta$ be a real number which is different from a negative integer. By the generalizing the result from Another sum involving binomial coefficients. and Some combinatorial identity. I derived the following identity: \begin{eqnarray} &&S^{a_1,a_2}_{n_1,n_2}(m) := \sum\limits_{i=0}^{m-1}\binom{i+a_1}{n_1} \binom{i+a_2}{n_2} \frac{1}{i+\theta} =\\ &&\sum\limits_{l=0}^{n_1+n_2} \frac{{\mathcal A}_l^{(a_1,a_2,n_1,n_2)}}{l} \left(\binom{\theta-1+m}{l} - \binom{\theta-1}{l}\right) +\\ &&\binom{a_1-\theta}{n_1} \binom{a_2-\theta}{n_2} \left[\psi(\theta+m) - \psi(\theta)\right] \end{eqnarray} Here the coefficients ${\mathcal A}_l^{(a_1,a_2,n_1,n_2)}$ read: \begin{eqnarray} &&{\mathcal A}_l^{(a_1,a_2,n_1,n_2)} = \sum\limits_{q_1=0}^{n_1} \binom{a_1-\theta}{n_1-q_1} \binom{a_2-\theta+q_1}{n_2-l+q_1} \binom{l}{q_1} 1_{l-n_2 \le q_1 \le l}\\ &&= \sum\limits_{q_2=0}^{n_2} \binom{a_1-\theta}{n_1-l+q_2} \binom{a_2-\theta+l-q_2}{n_2-q_2} \binom{l}{q_2} 1_{l-n_1 \le q_2 \le l} \end{eqnarray} and $\psi$ is the polygamma function of order zero. Note that \begin{equation} \left({\mathcal A}_l^{(a_1,a_2,n_1,0)},{\mathcal A}_l^{(a_1,a_2,0,n_2)}\right) = \left(\binom{a_1-\theta}{n_1-l}, \binom{a_2-\theta}{n_2-l} \right) \end{equation} and we retrieve the result from Another sum involving binomial coefficients. . The question is now what happens in case both $n_1$ and $n_2$ are allowed to be real.
If both $n_1$ and $n_2$ are real then the coefficients ${\mathcal A}_l$ can be expressed in terms of generalised hypergeometric series at unity as follows: \begin{eqnarray} {\mathcal A}_l^{(a_1,a_2,n_1,n_2)} = \binom{a_1-\theta}{n_1} \binom{a_2-\theta}{n_2-l} F_{3,2}\left[ \begin{array}{ccc} -l & -n_1 & 1+a_2-\theta \\ 1-l+n_2 & 1+a_1-n_1-\theta \end{array} ; 1 \right] \end{eqnarray} The identity on the top of this page holds except that the upper limit is replaced by infinity. In this way we obtain an infinite series. The question of convergence of that series still needs to be investigated.