Yes, it is well known, as it is the Vandermonde convolution in a disguised form
$$
\eqalign{
& \sum\limits_{k\, = \,0}^n {\left( \matrix{
n - k + a \cr
n - k \cr} \right)\left( \matrix{
k + b \cr
k \cr} \right)} = \cr
& = \sum\limits_{k\, = \,0}^n {\left( { - 1} \right)^{\,n - k} \left( \matrix{
- a - 1 \cr
n - k \cr} \right)\left( { - 1} \right)^{\,k} \left( \matrix{
- b - 1 \cr
k \cr} \right)} = ({\rm uppernegation}) \cr
& = \left( { - 1} \right)^{\,n} \sum\limits_{k\, = \,0}^n {\left( \matrix{
- a - 1 \cr
n - k \cr} \right)\left( \matrix{
- b - 1 \cr
k \cr} \right)} = \cr
& = \left( { - 1} \right)^{\,n} \left( \matrix{
- a - b - 2 \cr
n \cr} \right) = ({\rm Vandermonde}\;{\rm conv}{\rm .}) \cr
& = \left( \matrix{
n + a + b + 1 \cr
n \cr} \right)({\rm uppernegation}) \cr}
$$
With the "extended definition" of the binomial
$$
\left( \matrix{
x \cr
q \cr} \right) = x^{\,\underline {\,q\,} } /q!\;{\rm iff}\;0 \le {\rm integer}\,q,\;0\;{\rm otherwise}
$$
the result is valid also for $a,b$ real or even complex.