Is it possible to perform shell command substitution without using a subshell?

The choice of the title of your question is a bit confusing.

pushd/popd, a csh feature copied by bash and zsh, are a way to manage a stack of remembered directories.

pushd /some/dir

pushes the current working directory onto a stack, and then changes the current working directory (and then prints /some/dir followed by the content of that stack (space-separated).

popd

prints the content of the stack (again, space separated) and then changes to the top element of the stack and pops it from the stack.

(also beware that some directories will be represented there with their ~/x or ~user/x notation).

So if the stack currently has /a and /b, the current directory is /here and you're running:

 pushd /tmp/whatever
 popd

pushd will print /tmp/whatever /here /a /b and popd will output /here /a /b, not /tmp/whatever. That's independent of using command substitution or not. popd cannot be used to get the path of the previous directory, and in general its output cannot be post processed (see the $dirstack or $DIRSTACK array of some shells though for accessing the elements of that directory stack)

Maybe you want:

pushd "$(mktemp -d)" &&
popd &&
rmdir "$OLDPWD"

Or

cd "$(mktemp -d)" &&
cd - &&
rmdir "$OLDPWD"

Though, I'd use:

tmpdir=$(mktemp -d) || exit
(
  cd "$tmpdir" || exit # in a subshell 
  # do what you have to do in that tmpdir
)
rmdir "$tmpdir"

In any case, pushd "$(mktemp -d)" doesn't run pushd in a subshell. If it did, it couldn't change the working directory. That's mktemp that runs in a subshell. Since it is a separate command, it has to run in a separate process. It writes its output on a pipe, and the shell process reads it at the other end of the pipe.

ksh93 can avoid the separate process when the command is builtin, but even there, it's still a subshell (a different working environment) which this time is emulated rather than relying on the separate environment normally provided by forking. For example, in ksh93, a=0; echo "$(a=1; echo test)"; echo "$a", no fork is involved, but still echo "$a" outputs 0.

Here, if you want to store the output of mktemp in a variable, at the same time as you pass it to pushd, with zsh, you could do:

pushd ${tmpdir::="$(mktemp -d)"}

With other Bourne-like shells:

unset tmpdir
pushd "${tmpdir=$(mktemp -d)}"

Or to use the output of $(mktemp -d) several times without explicitly storing it in a variable, you could use zsh anonymous functions:

(){pushd ${1?} && cd - && rmdir $1} "$(mktemp -d)"

You could unlink the directory first, before leaving it:

rmdir "$(pwd -P)" && popd

or

rmdir "$(pwd -P)" && cd ..   # yes, .. is still usable

but note that pushd and popd are really tools for interactive shells, not for scripts (that's why they are so chatty; real scripting commands are silent when they succeed).