The following approach is based on the idea that, for positive integers $x,y$, the product $xy$ typically exceeds the sum $x+y$. We can apply this to show that in the problem of the question, either $a=b=c=d=2$ or else at least one of $a,b,c,d$ is 1; in this case it's easy to arrive at the only remaining solution $1,5;2,3$.
Suppose for two positive integers $x$ and $y$, that $x,y \ge k$ for some fixed $k \ge 2$. Then from $(x-k)(y-k) \ge 0$ we have, after expanding and rearranging, that $xy \ge k(x+y-k)$.
Now assume that each of $a,b,c,d$ is at least $k$. Of course we also use the assumptions $ab=c+d,cd=a+b$ of the question. We then have $ab \ge k(a+b-k)=k(cd-k)$, and also $cd \ge k(c+d-k)=k(ab-k)$. Putting these together we have
$ ab \ge k(k(ab-k)-k)=k^2ab-(k^3+k^2)$.
This inequality simplified is
$ab \le \frac{k^2}{k-1}.$
Now if $k=2$ here we obtain $ab \le 4$ which with the assumptions $a,b \ge 2$ leads to $a=b=2$, and (by symmetry or by the initial relations) also $c=d=2$.
If $k=3$ then we obtain $ab \le 9/2 = 4.5$, but this cannot hold since we're assuming $a,b \ge 3$ so that in fact $ab \ge 9$.
So by considering how products are typically larger than sums, we have shown that, except for the solution $a=b=c=d=2$, one of the values $a,b,c,d$ must be 1. Putting say $a=1$ into the two equations, one easily gets $a=1,b=5$ and that $c,d$ are $2,3$ insome order.