I came across this same binomial convolution in the following curious setting: consider the shift operator $S(a_n) = (a_{n+1})$
which maps
$(a_0, a_1, a_2, \ldots) \mapsto (a_1, a_2, a_3, \ldots)$.
It is easy to check that $S$ is a derivation of this convolution, that is:
$$ S ((a_n) \star (b_n)) = S (a_n) \star (b_n) + (a_n) \star S(b_n) $$
just by using Pascal's rule
$ {n+1 \choose k} = {n \choose k} + {n+1 \choose k-1} $.
This can be used to give a proof of the form of the general solution to a linear recurrence (homogeneous, with constant coefficients): Just repeat the same linear algebra one does to give a proof of the form of the general solution to a linear homogeneous differential equation with constant coefficients, exchanging the derivative operator D, functions and the exponential functions by the shift operator S, sequences and the geometrical sequences.
I have not seen this used elsewhere and I do not know if it has other applications other then the one sketched above.