What's the logic in exiting early on failure in blocks and subshells in Bash? [duplicate]

Question

In Bash blocks {} and subshells () the exit early doesn't work if there is an OR condition following it. Take for example

set -e
{ echo a; false; echo b; } || echo c 

prints

a
b

and

set -e
{ echo a; false; echo b; false;} || echo c 

prints

a
b
c

It seems to take the last executed command's exit code only. While this makes sense, given a semicolon instead of && is used, I'd expect the set -e to still make it exit on the first false and execute the error handling echo c code.

Using && instead of ; does make it work, but that makes it messy when having multiple line blocks. Adding in set -e at the start of the block/subshell also has no effect.

The reason this confuses me is because

set -e
{ echo a; false; echo b; }

prints

a

which means the exit-on-failure works when there's no || code following it. So I'd expect this to be the case with || code following it, executing it after the first failure in the block. Is there no way to achieve that without appending && after each line in the block?