We are interested in a sumproduct type estimate. Let $p$ be an odd prime, and let $A$ be the order $p1$ subgroup of $(\mathbb{Z}/p^2\mathbb{Z})^\times$. That is, let $A = \langle g^p \rangle$, where $(\mathbb{Z}/p^2\mathbb{Z})^\times = \langle g \rangle$. We seek an estimate of the size of the sum set $$A + A.$$ The standard references, e.g. Additive Combinatorics by Tao & Vu, do not discuss this problem in the setting of $\mathbb{Z}/p^2\mathbb{Z}$, but instead focus on analogous results in finite fields. Similar estimates address situations where $A + A$ is small (e.g., on the order of $KA$), whereas in our situation, the product set $A\cdot A$ is very small, and we wish to show that $A+A$ must be large. Do results exist that address this problem more directly?

HeathBrown and Konyagin stop just short of proving that $A+A\gg A^{3/2}$ in their paper on Gauss sums and Heilbronn's sum. (Here $A$ is the same subgroup as above; since $A\cup\{0\}=\{0^p,\ldots,(p1)^p\}$, we may write Heilbronn's sum as $H_p(a)=\sum_{n=0}^{p1}e(an^p/p^2)=1+\sum_{x\in A}e(ax/p^2)$.) By CauchySchwarz, we have $A\pm A\geq A^4/E_+(A)$, where $E_+(A)$ is the cardinality of the set of "additive quadruples" $\{(a,b,c,d)\in A^4\colon a+b=c+d\}$. Using Stepanov's method, HeathBrown and Konyagin show that $E_+(A)\ll (p1)^2+(p1)p^{3/2}$. (This follows from their lemma 4 and the argument at the bottom of page 7.) Combining the previous two results yields $A\pm A\gg A^{3/2}$, where the hidden constant is $1+o(1)$. In [http://arxiv.org/abs/math/0304217], Konyagin proves the same bound for multiplicative subgroups of $\mathbb{F}_p$ (see lemma 5). Shkredov gives the improved bound $E_+(A)\ll A^{42/17}(\logA)^{10/17}$ in [http://arxiv.org/abs/1208.6124v1], and a further improvement $E_+(A)\ll A^{22/9}(\logA)^{2/3}$ in [S]. Thus up to logs, our additive energy argument gives $A\pm A\gg A^{3/2+1/18}$. You can do better using an argument of Vyugin and Shkredov [VS] from [http://arxiv.org/abs/1102.1172]. Theorem 5.5 from [VS] shows that $R\pm R\gg R^{5/3}(\logR)^{1/2}$, where $R$ is a multiplicative subgroup of $\mathbb{F}_p$ of size at most $p^{1/2}$. The proof relies on the following results from [VS]: Lemma 2.3 and Corollary 2.7, which are true for any finite abelian group, and Corollary 5.1 and Lemma 5.4, whose analogs in [S] are Proposition 8 and Corollary 9. Thus we have $A\pm A\gg A^{5/3}(\logA)^{1/2}$. Finally, you can apply Proposition 8 from [S] to get a lower bound on A+A+A.
(Note that Shkredov uses $\circ$ to denote convolution with $B$; we're also using the convention that $B(x)$ is the indicator function of $B$.) If $Q_1=Q_2Q_3$, then the left hand side of the above equation is equal to $Q_2Q_3$. Choosing $Q_1=A+A+A$, $Q_2=A$ and $Q_3=(A+A)$, we get that



Yes. Bourgain has a sumproduct estimate for residues of a general modulus (although, the case of a composite modulus with few prime factors that covers your question was worked out prior by Bourgain and Chang to this). See: J. Bourgain, Sumproduct theorems and exponential sum bounds in residue classes for general modulus. C. R. Math. Acad. Sci. Paris 344 (2007), no. 6, 349–352 More precisely,
Taking $q=p^2$ and $A \sim p \sim q^{1/2}$ (which is the case in question) if we can prove that $\pi_{p}(A) > q^{\epsilon_{3}} $ for some absolute $\epsilon_3 >0$, then the sumproduct estimate (case II) will apply. Since the $A$ given is a multiplicative subgroup, we have that $A\cdot A = A$. Thus the sumproduce estimate will imply that $A+A > q^{\delta}A$. I now claim that $\pi_{p}(A) \geq p1 \sim q^{1/2}$. This requires a bit of elementary number theory: Claim: If $g$ is an integer that is a (multiplicative) generator mod $p^2$ then it is also a multiplicative generator mod $p$. Proof: Assume that $g$ isn't a generator mod $p$, then $g^{x} = 1 \mod p$ for some $x \leq p$. Rewrite this as $g^x = cp+1$. But now $g^{xp} \mod p^2 \equiv (cp+1)^p \mod p^2 \equiv 1 \mod p^2$, but since $Z_{p^2}$ has cardinality $p(p1)$ this would contradict that $g$ is a generator mod $p^2$.* Let $g$ be a generator mod $p^2$. By the claim, $g^b \mod p$ for $0 \leq b \leq p1 $ is a complete set of residues mod $p$. On the other hand $g^{pb} \mod p^2$ for $ 0 \leq b \leq p1 $ is a subset of the set $A$. By Fermat's little theorem $g^{pb} \equiv g^{b} \mod p $. Thus, $g^{pb} \mod p$ for $ 0 \leq b \leq p1 $ is a complete set of residues mod $p$. We conclude that $\pi_{p}(A) \geq p1 \sim q^{1/2}$. 

