$a+b=c \times d$ and $a\times b = c + d$

Define $r$, $s$ by $(x-a)(x-b)=x^2-rx+s$. Then $r=a+b=cd$ and $s=ab=c+d$, so $(x-c)(x-d)=x^2-sx+r$. Thus, the polynomials $x^2-rx+s$ and $x^2-sx+r$ both have only integer roots. Thus, $r^2-4s$ and $s^2-4r$ are both squares: $r^2-4s=u^2,s^2-4r=v^2$.

EDIT: Continuing, from $r^2-4s=u^2$ we get $4s=r^2-u^2\ge r^2-(r-2)^2=4r-4$, so $s\ge r-1$. By symmetry, $r\ge s-1$, so we have $r=s$ or $r=s\pm1$. If $r=s$ then $r^2-4r$ is a square, but $r^2-4r+4$ is a square, so we have two squares differing by $4$, which gives $r^2-4=0$, $r=0$ or $r=4$. These lead to the solutions $a=b=c=d=0$ and $a=b=c=d=2$.

$r=s+1$ implies $s^2-4s-4=(s-2)^2-8$ is a square. The only pair of squares differing by $8$ is $3^2-1^2$, and this leads to the $(1,5)$, $(2,3)$ solution (and its permutations) that we already have.

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.