The problem is the placement of the closing back-quote ` in the command:
export PATH=`echo ${PATH} | awk -v RS=: -v ORS=: '/SDE/ {next} {print}'` | sed 's/:*$//'
If you used the recommended $(...) notation, you'd see that this is equivalent to:
export PATH=$(echo ${PATH} | awk -v RS=: -v ORS=: '/SDE/ {next} {print}') | sed 's/:*$//'
which pipes the output of the export operation to sed, but export is silent.
Use:
export PATH=$(echo ${PATH} | awk -v RS=: -v ORS=: '/SDE/ {next} {print}' | sed 's/:*$//')
I have fixed the answer from which the erroneous command was copied verbatim. As tripleee notes in a comment, I'm not wholly convinced by the awk solution, but the question was 'what was wrong with the code' and the answer is 'where the back-quotes are placed'. The awk script does handle removing elements of a PATH at any position in the PATH; the sed script simply ensures there is no trailing : so that there is no implicit use of the current directory at the end of the PATH.
See also: How do I manipulate PATH elements in shell scripts and the clnpath script at How to keep from duplicating PATH variable in csh — the script is for POSIX-ish shells like the Bourne, Korn, Bash shells, despite the question's subject. One difference between clnpath and the notation used here is that clnpath only removes full pathnames; it does not attempt to do partial path element matching:
export PATH=$(clnpath $PATH /opt/SDE/bin)
if the path element to be removed was /opt/SDE/bin. Note that clnpath can be used to maintain LD_LIBRARY_PATH, CDPATH, MANPATH and any other path-like variable; so can the awk invocation, of course.
I note in passing that that the /SDE/ pattern in the awk script will remove /opt/USDER/bin; the slashes in the regex have nothing to do with slashes in the pathname.