Suppose $\mathbb{F}_3$ denotes the finite field $(F,\oplus,\odot)$, where $|F|=3$, and therefore $\oplus$ denotes addition modulo $3$, and $\odot$ denotes multiplication modulo $3$. Now, we wish to pick elements to fill the set $F$. We note that we may only pick three, since $F$ must satisfy $|F|=3$. The question in the title is whether we must pick elements such that $F$ is a complete residue system modulo $3$, or if we may pick elements such that $F$ is not? For example, since $0$, $1$, and $2$ are all possible residues modulo $3$, if we choose these elements to fill the set $F$, then $F$ is a complete residue system. Similarly, $6\mod3=0$, $10\mod3=1$, and $14\mod3=2$, so if we let $F=\lbrace6,10,14\rbrace$, then $F$ is, again, a complete residue system. We can construct the field for each of these cases as follows.
Case $1$: $F=\lbrace0,1,2\rbrace$
$$ \begin{array}{|c|c c c|} \hline \oplus & 0 & 1 & 2 \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 2 & 0 & 1 \\ \hline \end{array}~~~~~~\begin{array}{|c|c c c|} \hline \odot & 0 & 1 & 2 \\ \hline 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 \\ 2 & 0 & 2 & 1 \\ \hline \end{array} $$ Case $2$: $F=\lbrace6,10,14\rbrace$ $$ \begin{array}{|c|c c c|} \hline \oplus & 6 & 10 & 14 \\ \hline 6 & 6 & 10 & 14 \\ 10 & 10 & 14 & 6 \\ 14 & 14 & 6 & 10 \\ \hline \end{array}~~~~~~\begin{array}{|c|c c c|} \hline \odot & 6 & 10 & 14 \\ \hline 6 & 6 & 6 & 6 \\ 10 & 6 & 10 & 14 \\ 14 & 6 & 14 & 10 \\ \hline \end{array} $$
From the given tables, we note that in case $1$, the additive and multiplicative identities are $0$ and $1$, respectively. In addition, we see that $F$ is closed under $\oplus$, and $F\setminus\lbrace0\rbrace$ is closed under $\odot$, so $\mathbb{F}_3$ is a field when $F=\lbrace0,1,2\rbrace$. We may form the same line of reasoning for case $2$.
That being said, in both of these cases, $F$ is a complete residue system modulo $3$, as we have seen, but what if $F$ wasn't? If we pick $F=\lbrace0,1,4\rbrace$ (which we know is not a complete residue system since $4\equiv1\mod3$), then we know that, when constructing the table, we would run into the problem that $4\oplus4$ isn't mapped to any value in $F$, so it would make sense to conclude that $\mathbb{F}_3$ is not a field when $F$ is not a complete residue system, but is this true in general?
Any help is appreciated. Thank you.