Which is correct? Group Theory 
Unfortunately I noticed that all are wrong.
(A) Counterexample: $G=(1)$
(B), (C), (D) Counterexample: $G=\{1,-1\}$
Please help! Where did I go wrong?

You are right, and the question is wrong. (And you did not need to ask here to be sure; your counterexamples are convincing).
If it were your task to correct the question (and unfortunately readers of mathematical texts often find themselves in the uncomfortable position of having to correct apparent errors in what they are reading, and guessing the most probable oversight of the author) then you can choose from the additional hypotheses "$|G|>1$" (or $G$ is nontrivial) which makes (A) true, or "$2\nmid|G|$" (or $G$ is of odd order) which makes (D) true.
The first correction seems the most likely one. However it requires (very elementary) ring theory, not just group theory. If $G$ is a finite subgroup of the multiplicative group of a field (or domain) and $a\in G\setminus\{1\}$ then $a\sum_{g\in G} g=\sum_{g\in G}ag=\sum_{g\in G}g$ so $(a-1)\sum_{g\in G}g=0$ and since $a-1\neq0$ one has $\sum_{g\in G}g=0$.
If instead one adds the hypothesis $|G|$ is odd, then $\prod_{g\in G}g=1$. This follows from simple group theory and the fact that $G$ is Abelian (without which the product would make no sense). One can pair up mutually inverse elements in the product, which pairs contribute nothing, leaving only the involutions ($g^2=1$) single, so $\prod_{g\in G}g$ gives the same group restricted to the involutions only. In a group of odd order the identity is the only involution. (When instead $|G|$ is taken even then $\prod_{g\in G}g=-1$, since $-1$ is the unique element of order$~2$ in $G$, and even in $\Bbb C^*$.)
Added. The argument given for (A) with the assumption $|G|>1$ does not use commutativity of multiplication, so it holds for instance for finite subgroups of a skew field (a.k.a. division ring), which do not have to be cyclic. However, it does use the absence of zero divisors, and one might wonder if this can be avoided. The answer is it cannot, as the property sometimes fails in rings with zero divisors: the sum of both invertible elements in $\Bbb F_2[X]/(X^2)$ is $X$ rather than $0$.
