Let $a_k$ and $b_k$ be sequences of length $n$. Is there a closed form expression for the coefficients $c_k$ of $x^k$ for $k=0,...,n$ where,
$\displaystyle\prod_{k=1}^n(a_kx+b_k)=\displaystyle\sum_{j=0}^n c_jx^j$
I realize that the term "sequences" is rather broad. Specific cases where restrictions on said sequences are imposed are of interest.
This answer is based on the fact that $(a_k)$ and $(b_k)$ are recursive or iterative linear sequences, not for other types of sequences.
Let $(E_{\alpha,\beta})$ be the expression noted above, such that : $$ (E_{\alpha,\beta}) :\ \prod_{k=1}^n(a_k x + b_k)=\sum_{j=0}^n\left( c_j x^j \right) $$ where $ \alpha,\beta $ are functions such that : $$ \left\{ \begin{align} a_k &= \alpha(k)\\ b_k &= \beta(k) \end{align} \right. \ ,\forall k \leqslant n. $$
For such an expression, there isn't a definite answer; the sole expressions of my knowledge (from this page), and with research of others, where $c_j$ has a closed form, for all $j \leqslant n$, are :
$\begin{array}{ll}
\qquad\circ&\displaystyle (E_{(k-1),1}) :\ \prod_{k=1}^n(x + (k-1))=\sum_{j=0}^n\left( {n \brack j} x^j \right) \text{ where } c_j={n \brack j}, \forall j \leqslant n\\
\qquad\circ&\displaystyle (E_{(1-k),1}) :\ \prod_{k=1}^n(x + (1-k))=\sum_{j=0}^n\left( (-1)^{n-j}{n \brack j} x^j \right) \text{ where } c_j=(-1)^{n-j}{n \brack j}, \forall j \leqslant n
\end{array}$
and, if $(a_k)_{k \in [\![1,n]\!]}$ and $(b_k)_{k \in [\![1,n]\!]}$ are constant sequences of value $a_k=b_k=1$ for all $k \in [\![1,n]\!]$,
$\begin{array}{ll}
\qquad\circ&\displaystyle (E_{\text{const}}) :\ \prod_{k=1}^n(x + 1)=\sum_{j=0}^n\left( {n-1 \choose j} x^j \right) \text{ with : }\forall j \leqslant n,\, c_j={n-1 \choose j}\\
\end{array}$
You have
$$
\begin{gathered}
\prod\limits_{k = 1}^n {\left( {a_{\,k} x + b_{\,k} } \right)} = \prod\limits_{k = 1}^n {a_{\,k} \left( {x + \frac{{b_{\,k} }}
{{a_{\,k} }}} \right)} = \left( {\prod\limits_{k = 1}^n {a_{\,k} } } \right)\prod\limits_{k = 1}^n {\left( {x - \left( { - \frac{{b_{\,k} }}
{{a_{\,k} }}} \right)} \right)} = \hfill \\
= A\;\prod\limits_{k = 1}^n {\left( {x - r_{\,k} } \right)} \hfill \\
\end{gathered}
$$
so that's $A$ that multiplies a monic polynomial of degree $n$, having roots $r_k= - b_k/a_k$. Of course none of the $a$ terms shall be null, otherwise you just reduce the degree of the polynomial accordingly.
To express the polynomial it in the usual form
$$
A\,\sum\limits_{j = 0}^n {c_{\,j} \,x^{\,j} }
$$
is in what Vieta's formulas come in.
