We can prove the identity while we wait for a reference to appear.
We seek with $0\le c\le a$ and $0\le d\le b$
$${n+c\choose a} {n+d\choose b}
= \sum_{q=0}^{a+b} {a-c+d\choose q-c} {b-d+c\choose q-d}
{n+q\choose a+b}.$$
We get for the RHS
$$[z^{a+b}] \frac{1}{1-z}
\sum_{q\ge 0} z^q {a-c+d\choose a+d-q} {b-d+c\choose b+c-q}
{n+q\choose a+b}
\\ = [z^{a+b}] \frac{1}{1-z}
[w^{a+d}] (1+w)^{a-c+d} [v^{b+c}] (1+v)^{b-d+c}
\sum_{q\ge 0} z^q w^q v^q
{n+q\choose a+b}
\\ = [z^{a+b}] \frac{1}{1-z}
[w^{a+d}] (1+w)^{a-c+d} [v^{b+c}] (1+v)^{b-d+c}
[u^{a+b}] (1+u)^n
\\ \times \sum_{q\ge 0} z^q w^q v^q (1+u)^q
\\ = [z^{a+b}] \frac{1}{1-z}
[w^{a+d}] (1+w)^{a-c+d} [v^{b+c}] (1+v)^{b-d+c}
[u^{a+b}] (1+u)^n
\frac{1}{1-zwv(1+u)}
\\ = - [z^{a+b+1}] \frac{1}{1-z}
[w^{a+d+1}] (1+w)^{a-c+d} [v^{b+c}] (1+v)^{b-d+c}
\\ \times [u^{a+b}] (1+u)^{n-1}
\frac{1}{v-1/z/w/(1+u)}.$$
The contribution from $v$ is
$$\;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{b+c+1}} (1+v)^{b-d+c}
\frac{1}{v-1/z/w/(1+u)}.$$
Here we see that the residue at infinity is zero, so we may use minus
the residue at $v=1/z/w/(1+u)$ (residues sum to zero):
$$- (1+u)^{b+c+1} z^{b+c+1} w^{b+c+1}
\frac{(zw(1+u)+1)^{b-d+c}}{z^{b-d+c} w^{b-d+c} (1+u)^{b-d+c}}
\\ = - (1+u)^{d+1} z^{d+1} w^{d+1}
(zw(1+u)+1)^{b-d+c}.$$
Substitute into the remaining extractors,
$$[z^{a+b-d}] \frac{1}{1-z}
[w^a] (1+w)^{a-c+d} [u^{a+b}] (1+u)^{n+d}
(zw(1+u)+1)^{b-d+c}.$$
The contribution from $z$ is
$$\;\underset{z}{\mathrm{res}}\;
\frac{1}{z^{a+b-d+1}} \frac{1}{1-z}
(zw(1+u)+1)^{b-d+c}.$$
Using the the inequalities as stated in the preliminaries we find
again that the residue at infinity is zero, and we may use minus the
residue at $z=1.$ We also note that there is no pole at
$z=-1/w/(1+u).$ Collecting the remaining two extractors, we get
$$[w^a] (1+w)^{a-c+d} [u^{a+b}] (1+u)^{n+d}
(w(1+u)+1)^{b-d+c}
\\ = [w^a] (1+w)^{a-c+d} [u^{a+b}] (1+u)^{n+d}
(1+w+uw)^{b-d+c}
\\ = [w^a] (1+w)^{a-c+d} [u^{a+b}] (1+u)^{n+d}
\sum_{q=0}^{b-d+c} {b-d+c\choose q}
(1+w)^{b-d+c-q} u^q w^q
\\ = [u^{a+b}] (1+u)^{n+d}
\sum_{q=0}^{b-d+c} {b-d+c\choose q}
{a+b-q\choose a-q} u^q
\\ = \sum_{q=0}^{b-d+c} {b-d+c\choose q}
{a+b-q\choose a-q} {n+d\choose a+b-q}.$$
Here we have by construction that the rightmost two binomial
coefficients are zero when the lower index goes negative. This means
that we can replace the upper range by $a$. If $a\lt b-d+c$ we get for
$q\gt a$ that the second binomial coefficient is zero. If $b-d+c\lt a$
we get for $q\gt b-d+c$ that the first binomial coefficient is
zero. We thus have
$$\sum_{q=0}^a {b-d+c\choose q}
{a+b-q\choose a-q} {n+d\choose a+b-q}.$$
Now observe that
$${a+b-q\choose a-q} {n+d\choose a+b-q}
= \frac{(n+d)!}{(a-q)!\times b! \times (n+d+q-a-b)!}
\\ = {n+d\choose b} {n+d-b\choose a-q}.$$
We thus obtain
$${n+d\choose b} \sum_{q=0}^a
{b-d+c\choose q} {n+d-b\choose a-q}.$$
Here we may use Vandermonde which yields
$${n+d\choose b} {n+c\choose a}.$$
This is the claim. It is interesting to try this sum with the major
CAS.
