Submit AJAX form and update database then show all database results
Add a wrapper around the entire form, and return the entire form from your ajax callback. Then just build the rows reguarly in your form definition - The newly added rows will be shown since all the data is pulled from the database when the from is generated.

function clay_mod_blacklist_form($form, &amp;$form_state) {
  $form['#prefix'] = '&lt;div id="clay_mod_blacklist_form_wrapper"&gt;;
  $form['#suffix'] = '&lt;/div&gt;';

  // Add the rest of your form definition here

  $form['submit_button'] = array(
    '#type' =&gt; 'button',
    '#submit' =&gt; array('clay_mod_blacklist_form_submit'),
    '#ajax' =&gt; array(
      'callback' =&gt; 'clay_mod_blacklist_form_ajax',
      'wrapper' =&gt; 'clay_mod_blacklist_form_wrapper',
      'method' =&gt; 'after'
    ),
    '#value' =&gt; t('Add To Blacklist'),
  );

  $rows = db_select('blacklist', 't')
    -&gt;fields('t')
    -&gt;execute()
    -&gt;fetchAll();
  foreach($rows as $k =&gt; $v) {
    $form['results'][$k]['#markup'] = "&lt;div    class=\"vvvvvvvvv\"&gt;x{$v-&gt;name}y&lt;/div&gt;";
}

function clay_mod_blacklist_form_ajax(&amp;$form, &amp;$form_state)
{
  return $form;
}