NP-completeness of a Generalized Version of Subset Sum
Without loss of generality consider an instance $\langle S, t \rangle$ of subset sum where $S$ contains only positive integers and $t \ge 1$ (zeros can be dropped from $S$, and the case $t=0$ is trivial).
Now build a new instance $\langle T, t' \rangle$  of your generalized version of subset sum by choosing $T = \{ (t+1)x : x \in S \}$ and $t'=t(t+1)$.
If the elements of a subset $S' \subseteq S$ sum to $t$, then the elements of $\{ (t+1)x : x \in S' \} \subseteq T$ sum to $\sum_{x \in S'} (t+1)x = (t+1)\sum_{x \in S'} x =(t+1)t = t'$.
If there is a subset $T' \subseteq T$ of elements that can be arranged in an expression $E$ (that uses only addition, multiplication, and parentheses) that evaluates to $t'$, then $E$ uses no multiplication.
Indeed, if $E$ used at least one multiplication, it would evaluate to at least $(t+1)^2 &gt; (t+1)t = t'$ since each of the involved factors must be at least $(t+1)$.
As a consequence it must be that $t' = t(t+1) = \sum_{x \in T'} x$.
Let $S' = \{ \frac{x}{t+1} : x \in T' \} \subseteq S$. We have that $\sum_{x \in S'} x = \sum_{x \in T'} \frac{x}{t+1} = \frac{1}{t+1} \sum_{x \in T'} x = \frac{t'}{t+1} = t.$
This shows that your version of generalized subset sum is NP-complete (membership in NP is trivial).