Your question is interestingly tied to a generalization of Powers / Rising / Falling Factorials and Stirling Numbers.
Consider at first the monic polynomial
$$
p_{\,n} \left( {x,{\bf v}} \right) = \prod\limits_{0\, \le \;k\, \le \,n - 1} {\left( {x - v_{\,k} } \right)}
$$
then the product that you indicate corresponds to
$$
{\bf a} \star {\bf b} = p_{\,n} \left( {0, - {\bf a} - {\bf b}} \right)
$$
This tells us that, apart from simple translations and scaling, the properties
of this product cannot be put into a general frame, since there is not a simple relation
between polynomials with different roots.
Going a step further, consider the definition of the Rising Factorial.
$$
x^{\,\overline {\,n\,} } = x\left( {x + 1} \right) \cdots \left( {x + n - 1} \right) = \prod\limits_{0\, \le \;k\, \le \,n - 1} {\left( {x + k} \right)} = \prod\limits_{0\, \le \;k\, \le \,n - 1} {\left( {x + k} \right)}
$$
Then we can imagine to replace the "exponent" with a vector, and define
$$
x^{\,{\bf v}_{\,h} } = \left( {\matrix{
1 \cr
{x - v_{\,0} } \cr
{\left( {x - v_{\,0} } \right)\left( {x - v_{\,1} } \right)} \cr
\vdots \cr
{\left( {x - v_{\,0} } \right)\left( {x - v_{\,1} } \right) \cdots \left( {x - v_{\,h - 1} } \right)} \cr
} } \right) = \left\| {\;\prod\limits_{0\, \le \;k\, \le \,n - 1} {\left( {x - v_{\,k} } \right)} \quad \left| {\;0 \le n \le h} \right.\;} \right\|
$$
which means that a scalar raised to a vector is a vector, whose components are the mentioned polynomials in $x$ of
progressively increasing degree.
The vectors are indexed from $0$, and note that the last component of $\bf v$ does not come into play.
(these peculiar choices turn out to provide the best parallel with the "usual" definitions).
We have for instance
$$
x^{\,{\bf 0}_{\,h} } = \left( {\matrix{
1 \cr
x \cr
\vdots \cr
{x^{\,h} } \cr
} } \right)\quad \quad {\bf v} = \left( {\matrix{
0 \cr
1 \cr
\vdots \cr
h \cr
} } \right)\;\quad \Rightarrow \quad \;x^{\,{\bf v}_{\,h} } = \left( {\matrix{
{x^{\,\underline {\,0\,} } } \cr
{x^{\,\underline {\,1\,} } } \cr
\vdots \cr
{x^{\,\underline {\,h\,} } } \cr
} } \right)
$$
If we put
$$
p_{\,n} (z,{\bf v}) = \prod\limits_{0\, \le \,\,k\, \le \,n - 1} {\left( {z - v_{\,k} } \right)} = \sum\limits_{0\, \le \,\,k\, \le \,n} {a_{\,n,\,\,k} ({\bf v})\,z^{\,k} }
$$
then the Vieta's formulas will give us
$$
{\bf A}_{\,h} ({\bf v}) = \left\| {\;a_{n,\,m} \;} \right\|_{\,h} = \left\| {\;\left[ {m \le n} \right]\left( { - 1} \right)^{\,n - m} \sum\limits_{0\, \le \,k_{\,0} \, < \,k_{\,1} \, < \, \cdots \, < \,k_{\,n - m - 1} \, \le \,n - 1\;} {\left( {\prod\limits_{0\, \le \,j\, \le \,n - m - 1} {v_{\,k_{\,j} } } } \right)} \;} \right\|_{\,h}
$$
The inverse can be demonstrated to be
$$
{\bf A}_{\,h} ({\bf v})^{\, - \,{\bf 1}} = \left\| {\;\left[ {m \le n} \right]\sum\limits_{0\, \le \,k_{\,0} \, \le \,k_{\,1} \, \le \, \cdots \, \le \,k_{\,n - m - 1} \, \le \,m\;} {\left( {\prod\limits_{0\, \le \,j\, \le \,n - m - 1} {v_{\,k_{\,j} } } } \right)} \;} \right\|_{\,h}
$$
(note the difference between $<$ and $\le$ in the summation range).
So
$$
\;x^{\,{\bf v}_{\,h} } = {\bf A}_{\,h} ({\bf v})\,x^{\,{\bf 0}_{\,h} } \quad x^{\,{\bf 0}_{\,h} } = {\bf A}_{\,h} ({\bf v})^{\, - \,{\bf 1}} \,x^{\,{\bf v}_{\,h} }
$$
and it can be seen that ${\bf A}_{\,h} ({\bf v})$ and $ {\bf A}_{\,h} ({\bf v})^{\, - \,{\bf 1}} $ correspond respectively to the
(signed) Stirling N. of 1st kind and to those of the 2nd kind.
In conclusion, we are allowed to write
$$
{\bf a} * {\bf b}\quad \buildrel {{\rm one}\,{\rm element}\,{\rm of}} \over
\longrightarrow \quad 0^{\, - {\bf a} - {\bf b}}
$$
then
$$
0^{\, - {\bf a} - {\bf b}} = {\bf A}_{\,h} ( - {\bf a} - {\bf b})\;\,0^{\,{\bf 0}_{\,h} } = {\rm 1st}\,{\rm col}{\rm .}\,{\rm of}\;{\bf A}_{\,h} ( - {\bf a} - {\bf b})
$$
and from here we can establish various relations, like for instance
$$
0^{\, - {\bf a} - {\bf b}} = {\bf A}_{\,h} ( - {\bf a} - {\bf b})\,0^{\,{\bf 0}_{\,h} } = {\bf A}_{\,h} ( - {\bf a} - {\bf b})\,\;{\bf A}_{\,h} ( - {\bf a})^{\, - \,{\bf 1}} \;0^{\, - {\bf a}}
$$
But again, the last matrix product does not translate in general into a simple expression.