Whats the most elegant way to add two numbers that are Optional<BigDecimal>

I think the suggested answers of using streams or chains of methods on optionals are very clever, but perhaps so clever as to be obscure. The OP has modeled this as ordersTotal += newOrder with the exception that if both are empty, the result should be empty instead of zero. Maybe it would be reasonable to write the code so that it says that:

if (!ordersTotal.isPresent() && !newOrder.isPresent()) {
    result = Optional.empty();
} else {
    result = Optional.of(ordersTotal.orElse(ZERO).add(newOrder.orElse(ZERO)));
}

While this isn't the shortest, it clearly expresses exactly what the OP asked for.

Now I've assigned the computed value to result but the OP actually wanted to assign it back to ordersTotal. If we know both are empty, we can then skip the then-clause that assigns empty to ordersTotal. Doing that, and then inverting the condition gives something simpler:

if (ordersTotal.isPresent() || newOrder.isPresent()) {
    ordersTotal = Optional.of(ordersTotal.orElse(ZERO).add(newOrder.orElse(ZERO)));
}

Now, this tends to obscure the both-empty special case, which might not be a good idea. On the other hand, this says "add the values if either is non-empty" which might make a lot of sense for the application.

Not sure if you'll consider it more elegant, but here's one alternative:

ordersTotal = Optional.of(ordersTotal.orElse(BigDecimal.ZERO).add(newOrder.orElse(BigDecimal.ZERO)));

Another, based on @user140547's suggestion:

ordersTotal = Stream.of(ordersTotal, newOrder)
        .filter(Optional::isPresent)
        .map(Optional::get)
        .reduce(BigDecimal::add);

Note that the first version returns Optional.of(BigDecimal.ZERO) even when both optionals are empty, whereas the second will return Optional.empty() in such a case.