I'll start with the special case when A=B. Then we have the quadratic x^2-Ax+A. Then the sum of the roots is equal to the product of the roots.
Let the roots by y and z, then yz = y+z. Rearrange this into (y-1)*(z-1)=1. Solving over positive integers there is only one way to factor 1, then y-1=1 and z-1=1.
Then y=z=2 and then A=B=4. So the first solution to the problem is (A,B)=(4,4).
Continuing on, without loss of generality, assume A>B. Then x^2-Ax+B has two roots with a sum greater than their product.
Again let the roots by y and z. Then y+z > yz. Rearrange this into 1 > (y-1)*(z-1).
Over positive integers the only way for this to happen is when the left side is zero, which occurs when one of y or z equals 1. Then A=y+1 and B=y*1, and substituting to eliminate y leaves us with A=B+1.
At this point we have x^2-(B+1)x+B and x^2-Bx+(B+1). The first equation has roots of 1 and B. For the second equation to have integer solutions it is necessary for its discriminant to be a perfect square. This implies B^2-4(B+1) = B^2-4B-4 = (B-2)^2 - 8 = D^2 for some positive integers B and D.
This is a pair of perfect squares whose difference is only 8. There is only one such pair: 3^2 - 8 = 1^2
Therefore the only valid answer for B comes from B-2=3. Then B=5 and A=6.
So the only other solutions to the problem are (A,B)=(6,5) and its mirror (A,B)=(5,6).
Solution
