I add 9 to 9 which is 1100 in excess-3. I get 1 in carry and 1011, how can I show this [usefully] on the seven segments?

Displaying a conventional digit would blur any distinction in coding.
Hex and octal are even worse, sharing many symbols with digits of a base 10 representation but a different meaning.

I suggest mounting the 7-segement displays sideways to create a bit-wise display in the now vertical segments a, d, and g - 6 bits for two devices.

The sum of any two decimal digits is a single decimal digit (4 bits of excess-3), plus possibly a carry. That carry in excess-3 needs another 3 bits for a total of 7, but there are only 6. The two lower significant bits for the carry are identical for 0 (LHH) and 1 (HLL):
I suggest to display bit 0 of the carry digit with segment g of the more significant 7-segment display device, with the interpretation bit 0 and 1.
One could light the decimal point for error: bits 0 & 1 don't agree (a pity it is next to segment d).

I add 9 to 9 which is 1100 in excess-3. I get 1 in carry and 1011, how can I show this [usefully] on the seven segments?

Displaying a conventional digit would blur any distinction in coding.
Hex and octal are even worse, sharing many symbols with digits of a base 10 representation but a different meaning.

I suggest mounting the 7-segement displays sideways to create a bit-wise display in the now vertical segments a, d, and g - 6 bits for two devices.

The sum of any two decimal digits is a single decimal digit (4 bits of excess-3), plus possibly a carry. That carry in excess-3 needs another 3 bits for a total of 7, but there are only 6. The two lower significant bits for the carry are identical for 0 (LHH) and 1 (HLL):
I suggest to display bit 0 of the carry digit with segment g of the more significant 7-segment display device, with the interpretation bit 0 and 1.
One could light the decimal point for error: bits 0 & 1 don't agree. (A pity it isn't next to segment \$g\$ - light \$b\$&\$c\$ or \$e\$&\$f\$ instead?)