If you don't need to keep the order, and consider 45 and "45" to be the same:

function union_arrays (x, y) {
  var obj = {};
  for (var i = x.length-1; i >= 0; -- i)
     obj[x[i]] = x[i];
  for (var i = y.length-1; i >= 0; -- i)
     obj[y[i]] = y[i];
  var res = []
  for (var k in obj) {
    if (obj.hasOwnProperty(k))  // <-- optional
      res.push(obj[k]);
  }
  return res;
}

console.log(union_arrays([34,35,45,48,49], [44,55]));

You can do it using the underscore way:

// collectionUnion(*arrays, iteratee)
function collectionUnion() {
    var args = Array.prototype.slice.call(arguments);
    var it = args.pop();

    return _.uniq(_.flatten(args, true), it);
}

It just an improvment of the original function _.union(*arrays), adding an iteratee to work collection (array of object).

Here how to use it:

var result = collectionUnion(a, b, c, function (item) {
    return item.id;
});

The original function which is just working with array, looks like that:

_.union = function() {
  return _.uniq(flatten(arguments, true, true));
};

And in bonus a full example:

// collectionUnion(*arrays, iteratee)
function collectionUnion() {
    var args = Array.prototype.slice.call(arguments);
    var it = args.pop();

    return _.uniq(_.flatten(args, true), it);
}

var a = [{id: 0}, {id: 1}, {id: 2}];
var b = [{id: 2}, {id: 3}];
var c = [{id: 0}, {id: 1}, {id: 2}];

var result = collectionUnion(a, b, c, function (item) {
    return item.id;
});

console.log(result); // [ { id: 0 }, { id: 1 }, { id: 2 }, { id: 3 } ]