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alternative solution
we have
ab+c+d=ab-a-b+1+a+b+c+d-1=(a-1)(b-1)+a+b+c+d-1=3
bc+d+a=bc-b-c+1+a+b+c+d-1=(b-1)(c-1)+a+b+c+d-1=5
cd+a+b=cd-c-d+1+a+b+c+d-1=(c-1)(d-1)+a+b+c+d-1=2
da+d+a=da-d-a+1+a+b+c+d-1=(a-1)(d-1)+a+b+c+d-1=6
\iff
(a-1)(b-1)+a+b+c+d=4 \ \ \ \ (*_1)
(b-1)(c-1)+a+b+c+d=6 \ \ \ \ (*_2)
(c-1)(d-1)+a+b+c+d=3 \ \ \ \ (*_3)
(a-1)(d-1)+a+b+c+d=7 \ \ \ \ (*_4)
from this we have that
\begin{align}
(a-1)(b-1)(c-1)(d-1)
&=(4-(a+b+c+d))(3-(a+b+c+d))
\\&=(7-(a+b+c+d))(6-(a+b+c+d))
\\ \implies 12-7(a+b+c+d)+(a+b+c+d)^2&=42-13(a+b+c+d)+(a+b+c+d)^2
\\ \iff 6(a+b+c+d)&=30
\\ \iff a+b+c+d=5
\end{align}
this vastly simplifies the 4 initial equations to
(a-1)(b-1)=-1 \ \ \ \ (*_1)
(b-1)(c-1)=1 \ \ \ \ (*_2)
(c-1)(d-1)=-2 \ \ \ \ (*_3)
(a-1)(d-1)=2 \ \ \ \ (*_4)
(*_1)\implies a-1=\dfrac{1}{1-b}
(*_2)\implies b-1=\dfrac{1}{c-1}
(*_3)\implies c-1=\dfrac{2}{1-d}
(*_4)\implies d-1=\dfrac{2}{a-1}
we have (a-1)+(b-1)+(c-1)+(d-1)=5-4=1, and we can now express each of the bracketed terms in terms of solely any one of them, for each of the 4 possibilities, due to the above fraction relations. eg. we can say
(a-1)-\dfrac{1}{a-1}-\dfrac{2}{\frac{2}{a-1}}+\dfrac{2}{a-1}=1
\iff(a-1)-\dfrac{1}{a-1}-(a-1)+\dfrac{2}{a-1}=1
\iff\dfrac{1}{a-1}=1 \iff a=2
so now we have (b-1)+(c-1)+(d-1)=0. in a similar way,
(b-1)+\dfrac{1}{b-1}+\dfrac{2}{\frac{1}{1-b}}=0
\iff (b-1)+\dfrac{1}{b-1}+2(1-b)=0
\iff \dfrac{1}{b-1}-(b-1)=0
\iff (b-1)^2=1 \iff b=2,0
b=2 leads to a contradiction in (*_1) as we know a=2, so we can conclude b=0 is the only possibility.
now we have (c-1)+(d-1)=1, from which follows
-\dfrac{2}{d-1}+(d-1)=1
\iff (d-1)^2-(d-1)-2=0
\iff (d-1)=\dfrac{1±\sqrt{1+8}}{2}=\dfrac{1±3}{2}
\iff d=(1+2),(1-1)=3,0
d=0 leads to a contradiction in (*_4) as we know a=2, so we can conclude d=3 is the only solution.
then using (*_3) we can obtain c:
(*_3)\implies c-1=\dfrac{2}{1-3}\iff c=0
\therefore a=2,b=c=0,d=3
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always satisfying to plug in obtained values into the original equations lol
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