set -e not working as expected with conditional operators

[rpaufin1]

Consider the following script:

#!/bin/bash

f() {
        ls missing-file 2> /dev/null
        echo "test"
}

# 1
(set -e; f); ret=$?
if (( ret )); then
        echo "failed"
        exit 1
fi

# 2
(set -e; f) || {
        echo "failed"
        exit 1
}

Running the block labelled '1' prints "failed" and returns 1 as the exit code, 
while running the other block prints "test" and returns 0. However, the result 
should be the same.
If f() is modified like so, however:

f() {
        ls missing-file 2> /dev/null || return 1
        echo "test"
}

then it "works", but this also works without using set -e.

I thought this was an issue with subshells, but apparently the following script 
also doesn't work as expected:

#!/bin/bash

f() {
        ls missing-file 2> /dev/null
        echo "test"
}

set -e
f && :

Since f() is not ran in a subshell, the script should quit after the 'ls' 
command because of set -e, but "test" is still printed and the script exits 
with 0. To make this script work you can either remove the "&& :" part or the 
'echo' command from f(), but again, it doesn't seem like it's supposed to work 
this way.

# This works
f() {
        ls missing-file 2> /dev/null
}
set -e
f && :

# This also works
f() {
        ls missing-file 2> /dev/null
        echo "test"
}
set -e
f

What's going on here? I'm using the latest release, Bash 5.2.15.