Instead of providing you a "better" solution, I will try to give you some feedback the code.

Block indentation

function vowelsAndConsonants(s) {
var strConsonants = "";
var strVowels = "";
var i;
// ..

To work with code it is important that it is formatted in a readable way. There are some style guides out there for example from google

function vowelsAndConsonants(s) {
  var strConsonants = "";
  var strVowels = "";
  var i;
  // ..

Type Embedded in Name

var strConsonants = "";
var strVowels = "";

In this variable names the type is embedded in the name

Avoid placing types in method names; it's not only redundant, but it forces you to change the name if the type changes.

Array iteration and for...in

For your task is the order important:

Input string, output vowels and consonants to log, separately but in order

You can read on MDN about the for..in loop,that it do not guaranties a traversal in order:

Note: for...in should not be used to iterate over an Array where the index order is important. [...]

[...] iterating over an array may not visit elements in a consistent order. Therefore, it is better to use a for loop with a numeric index (or Array.prototype.forEach() or the for...of loop) when iterating over arrays where the order of access is important.

Make the Else-Statement Implicit

if (s.charAt(i) == "a" || 
    s.charAt(i) == "e" || 
    s.charAt(i) == "i" || 
    s.charAt(i) == "o" || 
    s.charAt(i) == "u") {
  strVowels += s.charAt(i);
} else if (s.charAt(i) != "a" || 
           s.charAt(i) != "e" || 
           s.charAt(i) != "i" || 
           s.charAt(i) != "o" || 
           s.charAt(i) != "u") {
  strConsonants += s.charAt(i);
}

Currently the if-else statement tries to express: If you are a, e, i, o, u do something, else if you are not from a, e, i, o, u do something.

This is semanticly the same as: If you are a, e, i, o, u do something, else do something.

Additional we can wrap the condition of the vowels into its own method to make the code more readable.

function isVowel(letter) {
    return letter === "a" || 
           letter === "e" || 
           letter === "i" || 
           letter === "o" || 
           letter === "u"
}

The if-statement could now look like

var letter = s.charAt(i)
if (isVowel(letter)) {
  strVowels += letter;
} else {
  strConsonants += letter;
}

String Concatenation..

strVowels += letter

Every time Strings get merged by + a new String gets created, because Strings are immutable, that means that for each concatenation new memory space gets allocated.

Better would be to use an array instead of a string an push into it.

vowels.push(letter)

Example Refactoring

function isVowel(letter) {
    return letter === "a" ||
           letter === "e" ||
           letter === "i" ||
           letter === "o" ||
           letter === "u"
}

function vowelsAndConsonants(s) {
    var consonants = [];
    var vowels = [];

    for (var letter of s) {
        if (isVowel(letter)) {
            vowels.push(letter)
        } else {
            consonants.push(letter)
        }
    }

    for (var vowel of vowels) {
        console.log(vowel);
    }

    for (var constant of consonants) {
        console.log(constant);
    }
}

and from here you can still use some methods like forEach or using the ternary operator to shorten an if-else

function isVowel(letter) {
    return letter === "a" ||
           letter === "e" ||
           letter === "i" ||
           letter === "o" ||
           letter === "u"
}

function print(x) {
    console.log(x)
}

function vowelsAndConsonants(s) {
    var consonants = [];
    var vowels = [];

    for (var letter of s) {
        isVowel(letter)
            ? vowels.push(letter)
            : consonants.push(letter)
    }

    vowels.forEach(print)
    consonants.forEach(print)
}

The shorter method can be achieved using Regex and it is also the fastest according to JSBEN.CH:

var str = "The quick brown fox jumps over a lazy dog";
var vowels = str.match(/[aeiou]/gi);
var consonants = str.match(/[^aeiou$]/gi);
    vowels.concat([''],consonants).forEach(function(k){
    	console.log(k);
    });

According to comments from others, the code can be further improved as follow:

const str = "The quick brown fox jumps over a lazy dog"; 
const vowels = str.match(/[aeiou]/gi); 
const consonants = str.match(/[^aeiou]/gi);   
vowels.concat([''],consonants).forEach(k => { console.log(k); } );