Observer pattern: Prevent multiple firings on changed internal property

I have an object Foo that has 2 public properties: A and B.

These properties can be changed from outside, as the application I'm working on is using a data-binding architecture. This is the front-end (client) code, but it's also communicating with a server to fetch data.

So my object is listening for changes to both A and B.

Any change in A or B is triggering the update() method, which triggers a request to the server.

I also have a piece of logic in my object Foo that says every time you change B, you have to reset A to its initial value.

My problem is, how do I deal with updating B without triggering the update() method twice?

Looking at the image above, the green path should always trigger an update. The red path ends up sometimes triggering 2 updates, because by changing B it changes A (if it's not the initial value), which always triggers an update.

I have to add that everything is asynchronous so the observers for my properties don't notify me immediately after changing objects. (design choice of the platform and language I'm working in – Polymer.JS). So this makes it harder to set a flag before changing B from A - I tried this but it didn't work, as the watcher method gets triggered after the chain of changes has executed.

As I come from the front-end world, with a background in graphic design, I don't have the knowledge necessary to spot the mistakes I might be doing, but I want to learn, so I'd like you to point me to reference material and/or books I can read about the correct applications of the observer pattern.

Is there a pattern that deals with this particular scenario and what are possible solutions to this?

If the observer pattern isn't the right choice for this, is there another pattern I could use internally that allows me to expose only the two properties A and B to the outside world?

A third option would be that I'm applying the right pattern, but the platform doesn't support what I'm trying to achieve. I'd like to know this so I can post the question on StackOverflow under the right tags.