Exclude the output from ssh and only log the error if found

The message you try to exclude is likely a ssh banner. This one is displayed on stderr, that's why the grep -v doesn't work in your script.

Just try to redirect stderr to stdin before the grep. It should work. I don't really understand your command, but it would look like this:

typeset -f | sshpass -e ssh -o StrictHostKeyChecking=no user@${IPADDRESS} "$(cat); IFERROR=$(checkscript); echo "$IFERROR"" 2>&1 | grep -v "AUTHORIZED" > ${SSHTEMPFILE}

PS: What's the purpose of $(cat) ?

I'm not sure what you're trying to do, but there's at least one obvious mistake in your script. You have

ssh "$(cat); …"$IFERROR"" | …

The fragment $IFERROR is not inside quotes, it is expanded by the local shell in order to build the SSH command that is executed. You want to execute echo "$IFERROR" on the remote side, so you need to quote the $ to prevent it from being expanded locally:

typeset -f | sshpass -e ssh -o StrictHostKeyChecking=no user@${IPADDRESS} "
  $(cat);
  IFERROR=$(checkscript);
  echo \"\$IFERROR\"" > ${SSHTEMPFILE} 2>&1

The checkscript command is also executed locally. If you meant for it to be executed remotely, change IFERROR=$(checkscript) to IFERROR=\$(checkscript).

You don't need to store the output of checkscript in a variable. You can, but it made your life more complicated for no benefit. Just get its output directly.

typeset -f |
sshpass -e ssh -o StrictHostKeyChecking=no user@${IPADDRESS} "$(cat); checkscript" > ${SSHTEMPFILE} 2>&1