set -e inside $( ) differs between bash and /bin/sh
In default-mode bash, command substitutions do not inherit errexit, so with set -e, x=$(false; echo ran) sets x=ran and the script continues (bash 5.2). Under bash --posix (also when bash is invoked as sh) and under dash, the substitution stops at false and the assignment fails, so the script exits with rc 1. The same script can therefore pass or abort depending on whether it is run as 'bash script' or through a #!/bin/sh line. To get one behaviour in bash scripts, use shopt -s inherit_errexit (bash >= 4.4), and keep risky substitutions in bare assignments, never inside local/export or command arguments, whose own status masks the failure.
Reproduced on Ubuntu 24.04 (bash 5.2.21, /bin/sh -> dash 0.5.12) with 'set -e; x=$(false; echo ran); echo continued': 'bash t.sh' rc=0 and prints x=ran. 'bash --posix t.sh', 'sh t.sh', 'dash t.sh' and bash run through a symlink named sh all give rc=1. 'shopt -s inherit_errexit' makes plain bash give rc=1 too. Something to watch for when combining it with local: under inherit_errexit, 'f(){ local x=$(false; echo ran); echo "x=$x"; }; f' prints 'x=' and returns rc=0. The substitution stops at false, but local's own status hides that, so with the shopt on you get an empty value and the script keeps going. That's why the post says to keep these in bare assignments.