As @manivel mentioned correctly, if you're using server-side processing, it's not possible.

With client-side processing, there are couple ways to do it.

SOLUTION 1: Submit form directly

The trick is to turn form elements from pages other than current into <input type="hidden"> before submitting the form.

var table = $('#example').DataTable();

// Handle form submission event
$('#frm-example').on('submit', function(e){
   var form = this;

   // Encode a set of form elements from all pages as an array of names and values
   var params = table.$('input,select,textarea').serializeArray();

   // Iterate over all form elements
   $.each(params, function(){
      // If element doesn't exist in DOM
      if(!$.contains(document, form[this.name])){
         // Create a hidden element
         $(form).append(
            $('<input>')
               .attr('type', 'hidden')
               .attr('name', this.name)
               .val(this.value)
         );
      }
   });
});

See this example for code and demonstration.

SOLUTION 2: Submit form via Ajax

Another solution is to submit the form via Ajax.

var table = $('#example').DataTable();

// Handle form submission event
$('#frm-example').on('submit', function(e){
   // Prevent actual form submission
   e.preventDefault();

   // Serialize form data
   var data = table.$('input,select,textarea').serialize();

   // Submit form data via Ajax
   $.ajax({
      url: '/echo/jsonp/',
      data: data,
      success: function(data){
         console.log('Server response', data);
      }
   });
});

See this example for code and demonstration.

NOTES

Please note that both solutions will work in client-side processing mode only.

LINKS

See jQuery DataTables: How to submit all pages form data for more details.