Clearly, one solution is
$$\begin{bmatrix} p & q & r \\ q & r & p \\ r & p & q \end{bmatrix}$$
or a $90^\circ$ rotation of the same. Each row and column clearly contains the same three numbers $\{p, q, r\}$. We now show that this is the only solution. Without loss of generality (scaling all entries by a positive constant), the number in the centre of the $3 \times 3$ matrix is $r = 1$. Let us denote the matrix corner values $a, b, c, d$. Then the matrix is:
$$\begin{bmatrix} a & x & b \\ ???? & 1 & ???? \\ c & y & d \end{bmatrix}.$$
Equating the products of the first row, the last row and the middle column gives:
$$abx = xy = cdy.$$
These immediately imply (as $x$ and $y$ are non-zero) that $y = ab$ and $x = cd$. A similar argument applies to the other mystery elements, and so the matrix is reconstructed from its corner elements as:
$$\begin{bmatrix} a & cd & b \\ bd & 1 & ac \\ c & ab & d \end{bmatrix}.$$

For the same argument with sums rather than products (as the central element is now zero), it follows that the middle edges are sums of the opposite corners and so:
$$\begin{align*} ab - 1 &= a + b - 2 \\ ac - 1 &= a + c - 2 \\ bd - 1 &= b + d - 2 \\ cd - 1 &= c + d - 2. \end{align*}$$
These factorise as:
$$0 = (a - 1)(b - 1) = (a - 1)(c - 1) = (b - 1)(d - 1) = (c - 1)(d - 1).$$
Thus, of any two adjacent corners, at least one must have a value of $1$. In particular, there must be one pair of opposite corners for which both values are $1$, since if two opposite corners are not both $1$, the other two corners must be $1$. Rotating if necessary, we may assume those corners are $b$ and $c$ which leads to the stated original form with $r = 1$.

The key observation is that two numbers are determined by their product and sum.
Lemma. If $a$, $b$, $u$, $v$ are real numbers that satisfy $a + b = u + v$ and $ab = uv$, then $\{a, b\} = \{u, v\}$.
*Proof.* Because
$$\begin{align*} (x - a)(x - b) &= x^2 - (a + b)x + ab \\ &= x^2 - (u + v)x + uv = (x - u)(x - v), \end{align*}$$
this quadratic has roots $a$, $b$ as well as roots $u$, $v$.

Considering the first row and third column of the matrix
$$\begin{bmatrix} a & b & c \\ \star & \star & u \\ \star & \star & v \end{bmatrix}$$
which have the same sum and the same product, the lemma implies that $\{u, v\} = \{a, b\}$. After swapping rows two and three if necessary, we even get $u = a$ and $v = b$:
$$\begin{bmatrix} a & b & c \\ \star & \star & a \\ \star & \star & b \end{bmatrix} .$$

For the same reason, the missing entries in the first column are $b, c$ and in the second column $a, c$ are missing. We can also work with the third column and the second or third row and obtain that the missing entries in the second row are $b, c$ and in the third row are $a, c$.
This already proves the statement of the problem. From the information collected, it is not hard to describe the possible shapes of the matrix.
As there is no $a$ missing in the first column, it now is clear that the bottom left corner must be occupied by $c$. In a similar way, we can now fill in the remaining numbers. As we may earlier have swapped the second and the third row, we now see that the matrix must have one of these shapes
$$\begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \quad \text{or} \quad \begin{bmatrix} a & b & c \\ c & a & b \\ b & c & a \end{bmatrix}.$$