Let $p$ be a prime. Let $S$ be a set of residues modulo $p$. Define $$S^2 = \{a \cdot b \mid a \in S, b \in S\}.$$
Question: How small can we make $|S|$ such that $\{0, 1, \cdots, p-2, p-1\} \in S^2$ ?
It seems that the optimal bound should be around $\sqrt{2p}$. This can either be seen from a probabilistic argument or by recognizing that if $|S| = k$ then $S^2$ can cover $\dbinom{k}2$ elements. I am able to get a bound of $ \sim 2 \sqrt{p}$ as follows. Let $g$ be a generator of $(\mathbb{Z}/p\mathbb{Z})^{\times}.$ Then we can have the following powers of $g$ in $S:$ $$S = \{ g, g^2, g^3, \cdots, g^{\sqrt{p}}, g^{2 \sqrt{p}}, g^{3 \sqrt{p}}, \cdots, g^{(\sqrt{p}-1) \sqrt{p}} \}.$$
Can we do better ?
This is possibly too long for a comment, but I think it is useful in looking at this problem. Let $G$ be a group. For $H,K\subseteq G$ define $$HK=\{hk\;|\; h\in H,\; k\in K\}$$ Then a good generalization of your question is
Notice that in your case $G=(\mathbb{Z}/p\mathbb{Z})^{\times}$, and really what you wanted was $S^2=\mathbb{Z}/p\mathbb{Z}$, not $S^2=G$. However as noted by the comments if $S^2=G$, then $(S\cup \{0\})^2=\mathbb{Z}/p\mathbb{Z}$, and if $S^2=\mathbb{Z}/p\mathbb{Z}$ then $0\in S$ and $(S\setminus \{0\})^2=G$. Thus the answer to your question is simply $+1$ to the more natural general question proposed with $G=(\mathbb{Z}/p\mathbb{Z})^{\times}$, because we just add $0$ to the minimal set.
Define $$\phi(G)=\min\{|S|\; |\; S\subseteq G, \; S^2=G\}$$ The first observation is quite obvious
So as noted by the comments since $(\mathbb{Z}/p\mathbb{Z})^{\times}\cong Z_{p-1}$ (where $Z_n$ is the cyclic group of order $n$), we can consider this problem in the slightly easier to understand additive group $\mathbb{Z}/p\mathbb{Z}$. And to understand the problem at hand we really only need to understand $\phi$ on cyclic groups. We then have two useful properties.
Some of the papers in the comments talk about difference basis. If $\Delta[G]$ is the smallest cardinality of a difference basis, then we have $\phi(G)\le 2\Delta[G]$. Using the results of those papers we then have $$\phi(G)\le \frac{8}{\sqrt{3}}\sqrt{|G|}\qquad\text{and}\qquad \phi(Z_n)\le 3\sqrt{n}$$ So for your question we have the bound $3\sqrt{p-1}$. However we can slightly improve this bound in specific cases. Note that
This is because the cosets of $H$ partition $G$, so pick out the set $R$ of representatives of the cosets, with $|R|=|G:H|=|G|/|H|$. Then $RH=G$, and so $(R\cup H)^2\supseteq RH=G$. Thus $\phi(G)\le |R\cup H|=|H|+|G|/|H|-1$. This gives us the following bound which is an improvement of the aforementioned ones in special cases
Since $G$ is CLT, there exists a subgroup of order $n\mid |G|$. Thus by the earlier property $\phi(G)\le n+nm=n(m+1)$.
For your question this is an improvement only when $p=n^2+1$, in which case we get the bound $2\sqrt{p-1}$. I think the best way to do better than this is look at the papers on difference bases and emulate them for your question.
Extra Stuff
The proof of course being if $S_1^2=H$ and $S_2^2=K$, then $(S_1\times S_2)^2=H\times K$. There's a chance that this is an equality, which I haven't been able to assess yet. And finally we notice that
Let $R$ be a fixed set of representatives of the cosets on $N$ with the group operation defined by $r_1,r_2\in R$, then $r_1r_2=r_3$ where $r_3$ is such that $r_1r_2N=r_3N$. Of course, $R\cong G/N$. Now let $R_0$ be such that $R_0^2=R$ and $|R_0|=\phi(R)=\phi(G/N)$, and let $S_0$ be such that $S_0^2=N$ and $|S_0|=\phi(N)$. Since $S_0\subseteq N\subseteq Z(G)$, we have $R_0S_0=S_0R_0$, and since the cosets of $N$ partition $G$, it follows that $$(R_0S_0)^2=R_0^2S_0^2=RN=\bigcup_{r\in R}rN=G$$ Thus $\phi(G)\le |R_0S_0|\le |R_0|\cdot |S_0|\le \phi(G/N)\phi(N)$.
This gives us the following property.
First it follows from Property 2 that if $|P|=p^{\alpha}$ then $\phi(P)\le \phi(Z_p)^{\alpha}$. This is because $p$-groups have a normal subgroup of order $p$ contained in their center, allowing us to inductively apply Property 2. Then since $G$ is nilpotent it is the direct product of its Sylow subgroups allowing us to apply this fact and Property 1 to get the result.