Try this method

1) Write your menu like this 

<!--ACTION-->
<record id="action_display" model="ir.actions.act_url">
     <field name="name">Menu Name</field> 
     <field name="url">/display_models</field> 
     <field name="target">self</field>
 </record>
<!--MENU -->
<record id="menu_action" model="ir.ui.menu">
    <field name="name">Menu Name</field> 
    <field name="sequence" eval="10"/> 
    <field name="action" ref="action_display"/>
</record>

2) Write Python code

from openerp import http 
from openerp.http import request

class YourClass(http.Controller):
    @http.route('/display_models', type='http', website=True) 
    def index_grid(self, **kw): 
             .....
        return request.render('your_module.your_template_id', {
                'model1':model1,'model2':model2,'model3':model3,'model4':model4})

3) Write Template

 <template id="your_module.your_template_id">
     <t t-call="web.menu"/><!--For menu bar-->
     <table>
        <tr>
            <td><t t-esc="model1.name"/><t t-esc="model1.partner_id"/> etc....</td>
            <td><t t-esc="model2.name"/><t t-esc="model2.partner_id"/> etc....</td>
        </tr>
        <tr><td><t t-esc="model3.name"/><t t-esc="model3.partner_id"/> etc....</td>
            <td><t t-esc="model4.name"/><t t-esc="model4.partner_id"/> etc....</td>
        </tr>
     </table>

     
 </template>

You could do it by creating a model0 with a many2one field for every model and related fields to all the fields of the models using the model name as a suffix for the related fields and the corresponding field path using the model many2one. That will cause you to have the model0 writing to the 4 other models fields but you could manage all like one into a single view form that you could divide into four areas as you like

yep, you can create one more model is `model0` then create 4 one2many fields to each your model

- one2many f1 to model1

- one2many f2 to model2

- one2many f3 to model3

- one2many f4 to model4

then design the form view for `model0` as you want