public void steppingNumber(int s, int e) {

As described in the problem statement, s is the starting number and e is the ending number. Why not call them start and end then? That way we don't have to refer to the problem statement to know what the variables mean.

I would also pick another name for the function. You are printing the stepping numbers, so call it printSteppingNumbers. As a general rule, functions should have verb names. They do things and their names should reflect that.

    String str = String.valueOf(s);

Why a string? In C, this would make sense as the difference between a letter and a number in C is trivial (char is an integer type). Java isn't as forgiving. Even if you do want a string, this is the wrong place to do it. Call the function with s (or start).

    if ( isSteppingNumber(start) ) {
        System.out.print(start + " ");
    }

I added braces ({}) around the then clause. This helps avoid a class of error where someone intends to add a new statement to the then clause but actually makes it run whether the if triggers or not. I also find it easier to read.

List<String> numbers = new ArrayList<>();

while(str.length() >= 3) { // get every 3-digit comma-separated number
    numbers.add(str.substring(str.length()-3));
    str = str.substring(0,str.length()-3);
}

This seems overly complicated. You don't need to generate a List here. It's sufficient to check each chunk as you go through.

while ( str.length() > 3 ) {
    if ( ! isSteppingSequence(str.substring(str.length()-3)) ) {
        return false;
    }
    str = str.substring(0,str.length()-3);
}

return isSteppingSequence(str);

You also don't need >= here. A simple > is better, as you always want there to be one chunk left to test outside the loop.

But we can do better if we keep it as a number at this point. First, we need a constant.

private final static int CHUNK_SIZE = 1000;

Now we can use it.

    while ( number > CHUNK_SIZE ) {
        if ( ! isSteppingSequence(number % CHUNK_SIZE) ) {
            return false;
        }

        number /= CHUNK_SIZE;
    }

    return isSteppingSequence(number);

Another constant.

private final static int BASE = 10;

Now we just need to define isSteppingSequence and we get:

private final static int BASE = 10;
private final static int CHUNK_SIZE = 1000;

private static void listSteppingNumbers(int start, int end) {
    while ( start <= end ) {
        if ( isSteppingNumber(start) ) {
            System.out.print(start + " ");
        }

        start++;
    }
}

private static boolean isSteppingNumber(int number) {
    if ( number < BASE ) {
        return false;
    }

    while ( number > CHUNK_SIZE ) {
        if ( ! isSteppingSequence(number % CHUNK_SIZE) ) {
            return false;
        }

        number /= CHUNK_SIZE;
    }

    // all the other sequences were stepping, so 
    // if this one is, it's a valid stepping number
    // otherwise not
    return isSteppingSequence(number);
}

private static boolean isSteppingSequence(int number) {
    int previous_digit = number % BASE;
    number /= BASE;
    while ( number > 0 ) {
        int current_digit = number % BASE;

        if ( Math.abs(previous_digit - current_digit) != 1 ) {
            return false;
        }

        previous_digit = current_digit;
        number /= BASE;
    }

    // if number was less than BASE, then it skipped the while loop
    // and all single digit numbers are valid stepping sequences
    // so true
    // if it made it through the while loop, 
    // all the adjacent digits differed by 1
    // so true
    return true;
}

So we now have a more readable version of your initial algorithm. Since it doesn't convert to a String and from characters, I suspect that it will be slightly faster.

You also ask if we can do better than stepping through every number in the range and testing to see if it is a stepping number. Certainly we can. We can generate the stepping numbers by composition. We don't have to search for them. I notice that this is described in more detail on Stack Overflow.