Why is the formal definition of Latin square equivalent with the informal? Informally, a latin square is a table where each element appears exactly once in each row and each column.
I know that this is probrably not an official definition of, however, it should somehow match the formal definition, right?
My question is: how does the table matches the definition about $a * x = b$ and $y * a = b$ having unique solutions for all $a, b$? In the table, each element is also unique in any row or column, but I am still not able to connect it.
Thank you for help.
The definition I work with is from Bergman´s Universal Algebra: Fundamentals and Selected Topics:

 A: In combinatorics, a Latin square can be defined as an $n \times n$ square matrix where each element (from a set of size $n$) appears exactly once in each row and each column.  The algebraic structure $\langle Q, \cdot \rangle$ is typically called a quasigroup.
If $Q$ is finite, we can write out the multiplication table for $Q$, and maybe it'd look something like this
$$
\begin{array}{c|cc|}
\cdot & 0 & 1 \\
\hline
0 & 0 & 1 \\
1 & 1 & 0 \\
\hline
\end{array}
$$
If we ignore the margins, we have matrix $L=L(i,j)$:
$$
\begin{array}{|cc|}
\hline
0 & 1 \\
1 & 0 \\
\hline
\end{array}.
$$
Importantly, the entry in cell $L(i,j) = i \cdot j$, by definition of $L$ (where $\cdot$ is the operation in $\langle Q, \cdot \rangle$).  Or, in other words $L(i,j)=k$ if and only if $i \cdot j = k$.
We can go from a Latin square $L$ to a quasigroup too (provided the symbol set $Q$ of $L$ is the same as its row and column indices), by defining $\cdot$ such that $i \cdot j = L(i,j)$ for all $i,j \in Q$.
Two distinct solutions to $a \cdot x = b$ in $\langle Q, \cdot \rangle$ would imply two distinct solutions to $L(a,x)=b$, thereby implying $b$ occurs twice in row $a$ in $L$, violating the Latin square property.  Conversely, if $b$ occurs twice in row $a$ in $L$, then there are two distinct solutions to $L(a,x)=b$, and thus two distinct solutions to $a \cdot x = b$ in $\langle Q, \cdot \rangle$, violating the quasigroup property.  (Likewise for columns and $x \cdot a = b$.)
Two caveats: (a) $\langle Q, \cdot \rangle$ might not be finite, in which case the combinatorial definition "appears exactly once in each row and each column" will need extending (what is a "row" when $Q$ is uncountable?).  (b) For the equivalence to work, we need the rows and columns of the Latin square to match the symbol set.
