您可以使用.wrapAll()并调用多个元素，因为它们不在一个父标记下

<details class="js-form-wrapper">
  <summary>summ</summary>
  <div>div 1</div>
  <div>div 2</div>
  <table class="responsive"></table>
  <div class="js-form-type-managed-file">child</div>
</details>

$(".js-form-wrapper table, .js-form-type-managed-file").wrapAll('<div class="my-wrapper"></div>');

$(document).ready(function() {
  $(".js-form-wrapper table, .js-form-type-managed-file").wrapAll('<div class="my-wrapper"></div>');
});

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<details class="js-form-wrapper">
  <summary>summary</summary>
  <div>div 1</div>
  <div>div 2</div>
  <table class="responsive"></table>
  <div class="js-form-type-managed-file">inspect here to see wrap</div>
</details>

可能有几种方法可以做到这一点，但一个简单的方法是：

$('table').next().addBack().wrapAll('<div class="my-wrapper">');

$('details table').next().addBack().wrapAll('<div class="my-wrapper">');

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<details class="js-form-wrapper">
  <summary></summary>
  <div></div>
  <div></div>
  <table class="responsive"></table>
  <div class="js-form-type-managed-file "></div>
</details>