We will use the fact that $n^2 \equiv 0 \pmod{3}$ if $3 \mid n$ and $n^2 \equiv 1 \pmod{3}$ otherwise. A consequence is that if the sum of the squares of $k$ integers is divisible by $3$ and $k$ is not divisible by $3$, then at least one of the $k$ integers is divisible by $3$.

Because $2025 = 3^4 \cdot 5^2$ is divisible by $3$, the above implies that at least one of $a, b, c, d$ must be divisible by $3$. The equation $ab = cd$ implies that a second of these integers must be divisible by $3$. But then two integers are left the squares of which add to a number which is divisible by $3$, so by the same reasoning, all four integers must be divisible by $3$.

Dividing now both sides of the equation $a^2 + b^2 + c^2 + d^2 = 2025$ by $9$ leaves us with the sum of four squares equal to $2025/9 = 225 = 3^2 \cdot 5^2$, which still is divisible by $3$. The same reasoning as above implies that the four integers must all be divisible by $3$, i.e. the original $a, b, c, d$ are all divisible by $9$.

Let $a = 9A$, $b = 9B$, $c = 9C$ and $d = 9D$, it then suffices to list solutions to
$$A^2 + B^2 + C^2 + D^2 = 25 \quad \text{and} \quad AB = CD.$$
By permuting, we can assume wlog that $A \le B \le D$ and $C \le D$.
We cannot have $D \ge 5$ or the sum on the LHS becomes too large. We also cannot have $D \le 2$ as the sum becomes too small. So $D$ can only be $3$ or $4$. If $D = 3$, the equation $AB = CD$ implies $3 \mid AB$ and, as $A \le B \le D$ we must have $B = 3$. But if $B = D = 3$ then $A^2 + C^2 = 7$ which does not have a solution in integers. The only remaining case, then, is $D = 4$, in which case $A^2 + B^2 + C^2 = 9$ whose only positive solution up to permutation is $(1, 2, 2)$. Because $AB = CD \ge 4$, we must have $A = B = 2$ and $C = 1$. Multiplying this solution by $9$ gives $(a, b, c, d) = (18, 18, 9, 36)$ as one solution to the original problem. Three more solutions are obtained by permutation, giving these four solutions
$(18, 18, 9, 36)$, $(18, 18, 36, 9)$, $(9, 36, 18, 18)$, $(36, 9, 18, 18)$.

We note that $ab = cd$ implies that
$$a^2 + b^2 + c^2 + d^2 = (a+b)^2 + (c-d)^2 = (a-b)^2 + (c+d)^2.$$
From the fact given at the start, it follows that if $x^2 + y^2$ is divisible by $3$, both integers, $x$ and $y$, must be divisible by $3$ as well. As $2025 = 3^4 \cdot 5^2$, this implies that each of the four numbers $a+b, a-b, c+d, c-d$ must be divisible by $3$, hence each of $a, b, c, d$ is divisible by $3$.

After dividing both sides of the equation $a^2 + b^2 + c^2 + d^2 = 2025$ by $9$ we can repeat this argument and obtain that $a, b, c, d$ are all divisible by $9$. Let $a = 9A$, $b = 9B$, $c = 9C$ and $d = 9D$. We then wish to solve
$$(A+B)^2 + (C-D)^2 = (A-B)^2 + (C+D)^2 = 25.$$
We assume $A \ge B > 0$ and $C \ge D > 0$ and consider permutations later.

There are two ways of writing $25$ as a sum of two squares: $25 = 3^2 + 4^2 = 5^2 + 0^2$. Noting that $A+B$ and $A-B$ have the same parity but cannot be equal, and similarly for $C, D$, we see that the only possibilities with $A, B, C, D$ satisfying $A \ge B > 0$ and $C \ge D > 0$ are
$$\begin{array}{llll} A+B=4 & C-D=3 & A-B=0 & C+D=5 & \text{or} \\ A+B=5 & C-D=0 & A-B=3 & C+D=4 \end{array}$$
which lead to $(A, B, C, D) = (2, 2, 4, 1)$ or $(A, B, C, D) = (4, 1, 2, 2)$. This gives the following four solutions $(a, b, c, d)$, the second and the third are obtained from the above by swapping $a$ and $b$, or $c$ and $d$:
$(18, 18, 36, 9)$ $(18, 18, 9, 36)$ $(36, 9, 18, 18)$ $(9, 36, 18, 18)$

The equation $ab = cd$ implies $a/c = d/b$. Let $m, n$ be coprime integers such that $a/c = d/b = m/n$. Then, there exist integers $r, s$ such that $a = rm$, $b = sn$, $c = rn$ and $d = sm$. The question becomes:
$$\begin{aligned} a^2 + b^2 + c^2 + d^2 &= (rm)^2 + (sn)^2 + (rn)^2 + (sm)^2 \\ &= (r^2 + s^2)(m^2 + n^2) = 2025 = 3^4 \cdot 5^2. \end{aligned}$$
From the observation made at the start (before solution 1) we see that for no $k \ge 0$ has the equation $x^2 + y^2 = 3^k$ a solution in positive integers. This shows that both, $r^2 + s^2$ and $m^2 + n^2$, must be divisible by $5$. Moreover, if an equation $m^2 + n^2 = 3^k \cdot 5$ with $k > 0$ has a solution, both, $m$ and $n$, must be divisible by $3$. This would contradict our assumption that $m$ and $n$ are coprime. Hence, we must have
$$r^2 + s^2 = 3^4 \cdot 5 \quad \text{and} \quad m^2 + n^2 = 5.$$
Assuming $r \le s$ and $m \le n$, and because $5 = 1^2 + 2^2$ is the only way to write $5$ as a sum of two squares, we have $(r, s, m, n) = (9, 18, 1, 2)$ and so $(a, b, c, d) = (9, 36, 18, 18)$.

Swapping $r$ with $s$ corresponds to interchanging $(a, b)$ with $(d, c)$; swapping $m$ and $n$ corresponds to interchanging $(a, b)$ with $(c, d)$. Doing both swaps interchanges $a$ with $b$ and $c$ with $d$. This gives the four solutions stated above.