PHP script isn't finding MySQL table row properly

try changing this

$check = "SELECT * FROM users WHERE email = '$_POST[email]'";

to

$check = "SELECT * FROM users WHERE email = '" . $_POST['email'] . "'";

also you want to change the if - fetch_row returns either null or an array (http://php.net/manual/en/mysqli-result.fetch-array.php) so what you want to do is if($data) instead of if($data == 1) as it never returns 1.

Modify your app.js

$(document).ready(function() {
    $("submit").click(function() {
        alert("This was a test.");
        var email = $("#email").val(); // This line...
        $.ajax({
            type: "POST",
            url: "main_login.php",
            data: {email : email}, // This line..
            success: function(data) {
                alert(data);
                $(".message").text(data);
            }
        });
    });
});

Modify your php:

<?php
    $conn = new mysqli('localhost', 'user', 'pass!', 'db');
    mysqli_set_charset($conn, 'utf8mb4');
    $check = "SELECT * FROM users WHERE email = '". $_POST['email'] ."'";
    $results = mysqli_query($conn, $check);
    $data = mysqli_fetch_array($results, MYSQLI_NUM);
    if($data == 1) {
        echo "Login successful";
    } else {
        echo "This account does not exist";
    }

    $conn->close();
?>

There are a few things you need to change. You can have a look here and here

The main problem lies in your app.js

/* attach a submit handler to the form */
$("#formID").submit(function(event) {

  /* stop form from submitting normally */
  event.preventDefault();

  /* get some values from elements on the page: */
  var $form = $( this ),
      url = $form.attr( 'action' );

  /* Send the data using post */
  var posting = $.post( url, { name: $('#emailID').val() } );

  /* Alerts the results */
  posting.done(function( data ) {
    alert('success');
  });
});

This is a better way of doing it where formIDis the id for your form and emailID is the id for your email input which you have to put into your html.

But a more simple way is not using a form, instead simple buttons and input fields with ids and then using

$("#buttonID").click(function() {
    var postData = $('#emailID').val();
    $.ajax({
        type: "POST",
        data : {'pd' : postData},
        url: "main_login.php",
        success: function(data) {
            alert(data);
            $(".message").text(data);
        }
    });

here emailID is the ID for the email input field and the buttonID if the ID for the button. Also, do include the jquery script into your html. Check for $_POST['pd'] in your php file.

your html can be as simple as

<input type="email" id="emailID"/>
<button id="buttonID">Click</button>

Note that your code has various security concerns.

EDIT:

main_login.php

<?php
   $conn = new mysqli('localhost', 'user', 'pass!', 'db');
   mysqli_set_charset($conn, 'utf8mb4');
   if(isset($_POST['pd'] && !empty($_POST['pd'])) {
          $check = "SELECT * FROM users WHERE email =" . $_POST[email];
          $results = mysqli_query($conn, $check);
          $data = mysqli_fetch_array($results, MYSQLI_NUM);
          $conn->close();
          if($data == 1) {
              echo "Login successful";
          } else {
              echo "This account does not exist";
          }
    }
    else { echo 'postdata not received'; }      
 ?>

Note that I have closed the db connection before sending the AJAX resopnse. Try keeping the response as the last line of the code while using AJAX as I have faced many problems in different situations if I keep some other code after AJAX response. But this is something you can ignore too if things work normal for you. Also do post the error you are getting while running the code.