Disclaimer: I'm typing this out as I'm thinking this through, so it's a bit disorganized. I'll go back and work on cleaning this up when I get the time.
Initially, looking at the rings:
The total of all the rings added together is 140, doublecounting the two numbers on the left and right between AD and CE, and triplecounting the three numbers in the middle. Since all the numbers add to 78, and we know of 9 and 10 in the middle, the two numbers on the sides + 2 times the remaining middle number must equal 24. There are only three combinations of numbers that fulfill this, with the first number corresponding to the one in the middle: (7,6,4), (6,5,7), (5,2,12), and (3,6,12).
Further testing of these combinations thins this down to:
(7,6,4) and (5,2,12), as if 6 is the center number, 7 and 5 have to be the remaining numbers in Ring E and the middle rows, which is impossible, and if 3 is the center number then regardless of if the number on the right is 6 or 12, then the remaining number in E will have to be 3 or 9, both of which are already used.
Looking at the rings:
We know that the two remaining numbers in B and in D must add up to 9. The combinations with remaining numbers that sum to this are (4,5), (3,6) and (2,7).
The remaining numbers in A must sum to 8, being either (3,5), (2,6), or (1,7). Since the number on the left in the third row must be 6, 4, 2 or 12, we can conclude that it must be 2 and 6, which means the two numbers in ring B must be 4 and 5, since a number from each combination is present in ring A. Since the center number can only be 5 or 4, we know the top row, center number must be 4 and the second row, third number is 5.
Now working from Ring E:
The number on the third row on the right must be either 12 or 2 from the prior calculation of the middle five numbers. If the number is 2, the remaining number in E would be 11, which is already used, so the number third row on the right must be 12 and the second number on the bottom row must be 1, and now the remaining number in Ring C must be 3.
We are remaining with the numbers 6, 2, and 7. The only combination that fulfills Ring D from before is (2,7) and since Ring A must contain 2 and 6, we can conclude that the first number in the top row is 6, the first number in the third row is 2, and the first number in the bottom row is 7.
Therefore we come to the final solution of: