A Cayley table of an finite group has to have every element exactly once in every row and exactly once in every column.
Proof that every element of a group has to be at most once in every row and at most once in every column:
Let $(G, \circ)$ be a group and $a, b, c, d \in G$ with:
(I) $a \circ b = d$
(II) $a \circ c = d \Leftrightarrow a = d \circ c^{-1}$
Then:
$\begin{align} (a \circ c) \circ (a \circ b) &= d \circ d \\ \Leftrightarrow d \circ (d \circ c^{-1} \circ b) &= d \circ d \\ \Leftrightarrow d \circ c^{-1} \circ b &= d\\ \Leftrightarrow c^{-1} \circ b &= e\\ \Leftrightarrow b &= c \end{align}$
As the group is finite, this also means it is exactly once in every row/column ($\forall a,b \in G: a \circ b = x$ with $x \in G$).
Now my question is:
Does a group with an infinite number of elements exist, that has not every element in every row/column of its Cayley table?
(I know that Cayley tables usually get used only for finite groups. But if set of the group has a countable number of elements, you can imagine a Cayley table. For example, $(\mathbb{Z}, +)$ has obviously every element in every row/column).