Handling Space in Filename for sed command

The problem you're seeing is that the space is being treated as a separator.

A simple example:

% echo a b | xargs ls
ls: cannot access a: No such file or directory
ls: cannot access b: No such file or directory

We can see ls tried to look for files a and b.

Instead we should tell xargs and grep to use a different separator; the best one is the NUL characater (\0). To do this you can use -Z with grep and -0 with xargs.

So your command becomes

grep -Z .... | xargs -0 ....

For example:

$ ls
en 1.json  en 2.json  en.json
$ grep -Z -El ABC *.json | xargs -0 sed -ibak -e 's/ABC/HHH/'
$ ls
en 1.json  en 1.jsonbak  en 2.json  en 2.jsonbak  en.json  en.jsonbak

We can see the 3 files got modified ("bak" files created).

If in doubt use a loop:

#!/usr/bin/env bash

while IFS= read -r file; do
    sed -ibak -e 's#ABC#123#g' -e 's#DFG#456#g' "$file"
done < <(grep -El 'ABC|DFG' *.json)

The above assumes your file names do not contain newlines.

You can use a JSON-aware tool such as jq to perform the replacements:

jq '."123" = ."ABC" | ."456" = ."DFG" | del(."ABC", ."DFG")'

Output

{
  "123": "/long_name",
  "456": "/{long_name}"
}

Applying this to all three files, and remembering to quote the filenames that contain spaces,

for file in en.json 'en 1.json' 'en 2.json'
do
    cp -p -- "$file" "$file.old"
    jq '."123" = ."ABC" | ."456" = ."DFG" | del(."ABC", ."DFG")'<"$file.old" >"$file" &&
        : rm -f -- "$file.old"
done

Remove the : from in front of rm to remove the saved original file(s).