The answer is in the affirmative; indeed,
If $S$ is a finite non-empty subset of any abelian group such that every element of $S$ is a sum of two other (possibly, equal to each other) elements, then $S$ has a non-empty, zero-sum subset.
For a complete proof, see this recent preprint by János Nagy, Péter Pach, and myself. The proof is a little too long to be presented here but at least, to indicate the general direction, here is our main lemma.
Lemma.
Suppose that $M$ is an integer square matrix of order $n$, representable as $A-I$ where $A$ has all its elements non-negative and all row sums equal to $2$, and $I$ is the identity matrix. Then there exist nonzero vectors $u,v\in\{0,1\}^n$ such that $M^Tu=v$; that is, there exists a system of rows of $M$ such that their sum is a nonzero, zero-one vector.
In fact, for our purposes it would suffice to have any nonzero vector $u$ satisfying $M^Tu\in\{0,1\}^n$; having $u$ itself to be zero-one is an extra (compare with Bill Thurston's answer above).
The proof of the main lemma is completely elementary, by induction on $n$.
Historically, the question seems to originate from a problem contributed by Dan Schwarz to the EGMO 2012 (European Girls’ Mathematical Olympiad):
Does there exist a set of integers such that every element of the set
is a sum of two other elements, while the set does not contain a
finite nonempty zero-sum subset?
The answer to this question is positive since the set here is allowed to be infinite.
(This is not really an answer, but I am new here so cannot comment).
This is a nice argument. I was wondering about lower bounds for $|AA+A|$ in the case when $A\subset{\mathbb{R}}$, trying to get an improvement on the exponent $3/2$ that you mentioned above. I'm sure there are others who could make the same confession, but perhaps this simple argument had been overlooked because of the assumption that $A$ is a set of reals. At least I can say that I hadn't seen the argument before.
It's unusual to get such a big improvement for a sum-product type problem in $\mathbb{Z}$ (as opposed to $\mathbb{R}$). It would be interesting if the argument could be tweaked to get some different optimal sum-product type results for sets of integers (or indeed for sufficiently well-separated sets as Brendan suggested). Maybe the dual problem of lower bounds for $|A(A+A)|$ should be considered for this approach, although I don't think it works quite so nicely.
By the way, a small correction; in the proof, you are right to say that the sets are distinct, but it seems to me that the crucial property that you are really using is that the sets are disjoint.
Best Answer
I could trace this question down to the paper of Erdős and Heilbronn "On the addition of residue classes mod $p$" (Acta Arith. 17 (1970), 227–229), where it is shown that for a prime $p$, if $A$ is a subset of the $p$-element group with $\lvert A\rvert>6\sqrt{3p}$, then the subset sum set of $A$ is the whole group.
It seems that the first to prove that $\lvert A\rvert>c\sqrt{\lvert G\rvert}$, with an absolute constant $c$, suffices to represent $0$ in any abelian group $G$, was Szemerédi ("On a conjecture of Erdős and Heilbronn", Acta Arith. 17 (1970), 227–229).
Some tightly related problems are considered in a paper of Eggleton and Erdős "Two combinatorial problems in group theory" (Acta Arith. 21 (1972), 111–116).
Three years later, Olson ("Sums of sets of group elements", Acta Arith. 28 (1975), 147–156) proved that if $A$ is a zero-sum-free set in a finite abelian group $G$, then $\lvert A\rvert<3\sqrt{\lvert G\rvert}$.
Hamidoune and Zémor ("On zero-free subset sums", Acta Arith. 78 (1996), 143–152) showed that for the group of prime order $p$ it suffices to have $\lvert A\rvert>(1+o(1))\sqrt{2p}$.
The exact result for the prime-order groups was obtained by Nguyen, Szemerédi, and Vu ("Subset sums modulo a prime", Acta Arith. 131 (2008), no. 4, 303–316).
There were several further improvements in the general case, with the current record $\lvert A\rvert<\sqrt{6\lvert G\rvert}$ due to Gao, Huang, Hui, Li, Liu, and Peng ("Sums of sets of abelian group elements", J. Number Theory, 208 (2020), 208–229). In fact, they prove a stronger result: if $A$ is a finite zero-sum-free subset of an abelian group, then $A$ spans at least as many as $1+\lvert A\rvert^2/6$ pairwise distinct subset sums.
Two other important papers on this subject are mentioned in the answer of Le p'tit bonhomme.