How to use the S-output-modifier with Unix/Linux command ps?

I am not sure, but it might be that ps is somewhat buggy in this respect. Try this here:

$ ps p 12104 k timePID TTY      STAT   TIME COMMAND12104 ?        Ss    16:17 /usr/sbin/apache2 -k start$ ps p 12104 k time SPID TTY      STAT   TIME COMMAND12104 ?        Ss   143:16 /usr/sbin/apache2 -k start

This is using the BSD options for ps. It works on my machine, however you get an extra header row and extra columns. I would cut them away using tr and cut:

$ ps p 12104 k time S | tail -n 1 | tr -s '[:space:]' | cut -d ' ' -f 4143:39$ ps p 12104 k time | tail -n 1 | tr -s '[:space:]' | cut -d ' ' -f 416:17

On MacOS X (10.7, Lion) the manual page says:

-S Change the way the process time is calculated by summing all exited children to their parent process.

So, I was able to get output using:

$ ps -S -o time,etime,pid -p 305     TIME     ELAPSED   PID  0:00.12 01-18:31:07   305$

However, that output was not really any different from when the '-S' option was omitted.

I tried:

$ ps -S -o time,etime,pid -p 305     TIME     ELAPSED   PID  0:00.14 01-18:43:59   305$ time dd if=/dev/zero of=/dev/null bs=1m count=100k102400+0 records in102400+0 records out107374182400 bytes transferred in 15.374440 secs (6983941055 bytes/sec)real    0m15.379suser    0m0.056ssys     0m15.034s$ ps -S -o time,etime,pid -p 305     TIME     ELAPSED   PID  0:00.14 01-18:44:15   305$

As you can see, the 15 seconds of system time spent copying /dev/zero to /dev/null did not get included in the summary.

At this stage, the only way of working out what the '-S' option does, if anything, is to look at the source. You could look for sumrusage in the FreeBSD version, for example, at FreeBSD.