[proofplan]
We construct the correspondence in both directions and verify that each construction lands in the claimed class. A quasigroup operation gives a Latin square because left and right translations are bijections, so every row and every column contains each symbol exactly once. Conversely, a Latin square gives a quasigroup operation because the Latin condition gives unique solutions in the appropriate row or column. Finally, the two constructions are inverse to each other by their definitions.
[/proofplan]
[step:Send a quasigroup operation to its multiplication table]
Let $\circ: Q \times Q \to Q$ be a binary operation such that $(Q,\circ)$ is a quasigroup. Define the associated table to be the map
\begin{align*}
L_\circ &: Q \times Q \to Q,\qquad L_\circ(x,y) = x \circ y.
\end{align*}
We prove that $L_\circ$ is a Latin square with row set $Q$, column set $Q$, and symbol set $Q$.
Fix $x \in Q$. Define the left translation by $x$ to be the map
\begin{align*}
\lambda_x &: Q \to Q,\qquad \lambda_x(y) = x \circ y.
\end{align*}
Since $(Q,\circ)$ is a quasigroup, for every $b \in Q$ there exists a unique $y \in Q$ such that $x \circ y=b$. Therefore $\lambda_x$ is bijective. Hence the row of $L_\circ$ indexed by $x$ contains every symbol $b \in Q$ exactly once.
Fix $y \in Q$. Define the right translation by $y$ to be the map
\begin{align*}
\rho_y &: Q \to Q,\qquad \rho_y(x) = x \circ y.
\end{align*}
Since $(Q,\circ)$ is a quasigroup, for every $b \in Q$ there exists a unique $x \in Q$ such that $x \circ y=b$. Therefore $\rho_y$ is bijective. Hence the column of $L_\circ$ indexed by $y$ contains every symbol $b \in Q$ exactly once.
Thus $L_\circ$ is a Latin square indexed by $Q$ in its rows, columns, and symbols.
[guided]
Let $\circ: Q \times Q \to Q$ be a binary operation making $(Q,\circ)$ a quasigroup. We build its multiplication table by defining the map
\begin{align*}
L_\circ &: Q \times Q \to Q,\qquad L_\circ(x,y) = x \circ y.
\end{align*}
The input pair $(x,y)$ records the row $x$ and column $y$, and the value $L_\circ(x,y)$ is the symbol written in that cell.
To prove that this table is Latin, we must check two conditions: each row contains every symbol exactly once, and each column contains every symbol exactly once.
First fix a row index $x \in Q$. The row indexed by $x$ is controlled by the map
\begin{align*}
\lambda_x &: Q \to Q,\qquad \lambda_x(y) = x \circ y.
\end{align*}
A symbol $b \in Q$ occurs in this row precisely at the columns $y \in Q$ satisfying $x \circ y=b$. The quasigroup property says that for the equation $x \circ y=b$, with $x$ and $b$ fixed, there exists a unique solution $y \in Q$. Therefore each symbol $b \in Q$ appears in the row indexed by $x$ exactly once. Equivalently, $\lambda_x$ is bijective.
Now fix a column index $y \in Q$. The column indexed by $y$ is controlled by the map
\begin{align*}
\rho_y &: Q \to Q,\qquad \rho_y(x) = x \circ y.
\end{align*}
A symbol $b \in Q$ occurs in this column precisely at the rows $x \in Q$ satisfying $x \circ y=b$. The quasigroup property says that for the equation $x \circ y=b$, with $y$ and $b$ fixed, there exists a unique solution $x \in Q$. Therefore each symbol $b \in Q$ appears in the column indexed by $y$ exactly once. Equivalently, $\rho_y$ is bijective.
Since every row and every column contains each symbol of $Q$ exactly once, the table $L_\circ$ is a Latin square with row set $Q$, column set $Q$, and symbol set $Q$.
[/guided]
[/step]
[step:Send a Latin square to its induced operation]
Let $L: Q \times Q \to Q$ be a Latin square with row set $Q$, column set $Q$, and symbol set $Q$. Define a binary operation
\begin{align*}
\circ_L &: Q \times Q \to Q,\qquad x \circ_L y = L(x,y).
\end{align*}
We prove that $(Q,\circ_L)$ is a quasigroup.
Let $a,b \in Q$. Since row $a$ of $L$ contains each symbol of $Q$ exactly once, there exists a unique $x \in Q$ such that $L(a,x)=b$. By the definition of $\circ_L$, this is exactly the unique solution of
\begin{align*}
a \circ_L x=b.
\end{align*}
Similarly, since column $a$ of $L$ contains each symbol of $Q$ exactly once, there exists a unique $y \in Q$ such that $L(y,a)=b$. By the definition of $\circ_L$, this is exactly the unique solution of
\begin{align*}
y \circ_L a=b.
\end{align*}
Thus, for every $a,b \in Q$, both equations $a\circ_L x=b$ and $y\circ_L a=b$ have unique solutions in $Q$. Therefore $(Q,\circ_L)$ is a quasigroup.
[/step]
[step:Verify that the two constructions are inverse maps]
Let $\circ: Q \times Q \to Q$ be a quasigroup operation. The operation induced by its table $L_\circ$ is the map
\begin{align*}
\circ_{L_\circ} &: Q \times Q \to Q,\qquad x \circ_{L_\circ} y = L_\circ(x,y).
\end{align*}
For every $(x,y) \in Q \times Q$, the definition of $L_\circ$ gives
\begin{align*}
x \circ_{L_\circ} y = L_\circ(x,y) = x \circ y.
\end{align*}
Hence $\circ_{L_\circ}=\circ$.
Conversely, let $L: Q \times Q \to Q$ be a Latin square. The table associated to the induced operation $\circ_L$ is the map
\begin{align*}
L_{\circ_L} &: Q \times Q \to Q,\qquad L_{\circ_L}(x,y) = x \circ_L y.
\end{align*}
For every $(x,y) \in Q \times Q$, the definition of $\circ_L$ gives
\begin{align*}
L_{\circ_L}(x,y) = x \circ_L y = L(x,y).
\end{align*}
Hence $L_{\circ_L}=L$.
Therefore the two constructions are mutually inverse, so they define a bijection between quasigroup operations on $Q$ and Latin squares indexed by $Q$.
[/step]
