Newlines are handled specially by bash, regardless of the value of IFS. A backslash before a newline causes the newline to be ignored. Has nothing to do with word splitting and would occur even if IFS were set to some custom value.

From LESS=+/^QUOTING man bash:

 A non-quoted backslash (\) is the escape character. It preserves the literal value of the next character that follows, with the exception of <newline>. If a \<newline> pair appears, and the backslash is not itself quoted, the \<newline> is treated as a line continuation (that is, it is removed from the input stream and effectively ignored). 

You can see the word splitting on newline occur another way: By stuffing a newline into a variable, and then echoing the variable without quoting it.

$ myvar=a$'\n'b$'\n'c $ echo "$myvar" a b c $ echo $myvar a b c $ 

Direct answer

The unquoted command: echo a b c, will be split on (metacharacter space) no matter what IFS is or is not. The split arguments to echo: a, b and c will be printed with spaces because echo is defined as this:

LESS=+/'^ *echo \[-neE\] \[arg ...\]' man bash 

Output the args, separated by spaces, followed by a newline.

Also no $IFS involved.

So, what you quoted has no relevance here. The key part is:

... splits the results of the other expansions into words ...

You must have an expansion before $IFS is used.
There is not any expansion in a b c.

If you must have an effect of $IFS, use an unquoted expansion:

$ var='a b c' $ echo $var a b c 

But that still does not use $IFS for output, only for splitting.

IFS in output.

Where $IFS does have relevance (for output) is in $*.

$ set -- a b c $ IFS=$'\t' $ printf '%s\n' "$*" | od -An -vtx1c 61 09 62 09 63 0a a \t b \t c \n 

Is that what you need?

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Other issues.

Keyboard Tab

when I hit the key Tab, there is no response

Then press together CtrlV, release, and then press Tab.

But without quotes you will not be able to assign a tab to a variable. Test:

$ var=a b ### The space represents a tab as above. bash: b: command not found 

You will need:

$ var="a b" ### The space represents a tab as above. 

Double quotes at the very least.

Quote IFS

nothing comes out of $IFS:

$ echo $IFS 

Nothing comes out of $IFS because you did not quote $IFS. Try:

<user>$ echo "$IFS" | sed -n l \t$ $ <user>$ 

Which has one new line from $IFS and one from the command echo.
Better change to printf:

<user>$ printf '%s' "$IFS" | sed -n l \t$ <user>$ 

And use od:

$ printf '%s' "$IFS" | od -An -vtx1c 20 09 0a \t \n 

As we already have a way to "see" what characters are contained in an string.
We may as well use that to "see" what is inside a couple of strings:

$ var='a b c' $ echo "$var" | od -An -vtx1c 61 20 62 20 63 0a a b c \n 

That should not be any surprise. As this (with single quotes) should not be:

$ var='a\ > b\ > c' $ echo "$var" | od -An -vtx1c 61 5c 0a 62 5c 0a 63 0a a \ \n b \ \n c \n 

Exactly what was typed. If the var is properly quoted, the backslash appear correctly.

Quoting

But quoting with double quotes may change some characters:

$ var="a\ > b\ > c" $ echo "$var" | od -An -vtx1c 61 62 63 0a a b c \n 

But what you wrote has no quotes on setting the var nor using it on echo.
That is a no-no rule in shells. Please Quote correctly.

To replace a space with a tab character using $IFS you can use the argument array:

unset IFS set a b c IFS=$(printf \\t) printf "%s\n" "$*" 

Understand that $IFS is about splitting - it contains a list of characters the shell uses to split unquoted expansions in list contexts. You can't really use it to replace characters except in the special case of the $* special shell parameter. In all other cases its use will amount to nullifying and delimiting certain portions of unquoted shell expansions.