I will prove the following stronger result from A.K. Gupta, Generalized hidden hexagon squares, Fibonacci Quarterly, 1974: $$ \binom{n}{m-r} \binom{n-s}{m} \binom{n+r}{m+s} = \binom{n-s}{m-r} \binom{n}{m+s} \binom{n+r}{m} $$
Proof:
Let me write $\binom{a}{b,c}$ for the number of ways to choose two disjoint sets, of sizes $b,c$ respectively, from a set of size $a$. There are a few ways to do this: you could first pick $b$ of them, and then pick $c$ from the remaining, or you could pick $b+c$, and then decide how to divy them up, or... Thus, we have the following equalities: $$ \binom{n}{m-r,s} = \binom{n}{m-r}\binom{n-(m-r)}{s} = \binom{n-s}{m-r} \binom{n}{s} $$ $$ \binom{n}{m,s} = \binom{n-s}{m} \binom{n}{s} = \binom{n}{m+s} \binom{m+s}{s} $$ $$ \binom{n+r}{m,s} = \binom{n+r}{m+s} \binom{m+s}{s} = \binom{n+r}{m} \binom{(n+r)-m}{s} $$ Multiplying these equations together and canceling like terms gives the desired equality.
This does not provide a combinatorial interpretation when both sides are $0$ --- indeed, it applies only when all of $s$, $m$, $n-s$, and $n+r-m-s$ are nonnegative integers, as in other situations I had asked to divide by $0$.
I will not try to improve this, as you asked for a combinatorial proof, and I won't try to give combinatorial interpretation to "the number of ways to pick negatively many objects form a set with negative size".
In words, I start with sets of size $n$, $n$, and $n+r$, and remove a set of size $s$ from each, and then remove from what remains sets of size $m-r$, $m$, and $m$; I count the ways to do this in various ways, and notice one coincidence, that $n-(m-r) = (n+r) - m$.