WLOG $a\geq b$ and $c\geq d$ and $d\geq b$ If $b\geq 2$ then $d\geq 2$ so $a+b+c+d-3\geq 2a+2c\Leftrightarrow b+d-3\geq a+c\geq b+d$,contradiction So $b=1$ . $a+1+c+d-3=a+cd\Leftrightarrow c+d-2=cd\Leftrightarrow $ $\Leftrightarrow d-2=c(d-1)\Rightarrow d-1|d-2\overset{d>0}{\Leftrightarrow} d=2 ,c=0$ ,contradiction since $c>0$ So there is no solutions.

The problem in actual Romanian TST said $a+b+c+d-3=ab=cd$, not $a+b+c+d-3=ab+cd$. That makes the problem a little more interesting and, at least, produces a non-empty answer.