### Is there an existing issue for this?

- [X] I have searched the existing issues and read the documentation

### Question

I am using serverless framework with localstack and I am unable to define an `AWS::OpenSearchService::Domain` resource. I guess I'll have to use `AWS::Elasticsearch::Domain` instead but it is deprecated. Are there any plans to add `AWS::OpenSearchService::Domain` resource soon? 

### Anything else?

_No response_
bucket_name = resource_props.get("BucketName")
            bucket_name = resolve_refs_recursively(stack, bucket_name)
            if attribute == "Arn":
                return aws_stack.s3_bucket_arn(bucket_name)
            return bucket_name
    elif resource_type == "Elasticsearch::Domain":
        if attribute == "DomainEndpoint":
            domain_status = resource_state.get("DomainStatus", {})
            result = domain_status.get("Endpoint")
            if result:
                return result
        if attribute in ["Arn", "DomainArn"]:
            domain_name = resource_props.get("DomainName") or resource_state.get("DomainName")
            return aws_stack.es_domain_arn(domain_name)
    elif resource_type == "StepFunctions::StateMachine":
        if is_ref_attr_or_arn:
            return resource_state["stateMachineArn"]
    elif resource_type == "SNS::Topic":
        if is_ref_attribute and resource_state.get("TopicArn"):
            topic_arn = resource_state.get("TopicArn")
            return resolve_refs_recursively(stack, topic_arn)
    elif resource_type == "SQS::Queue":
        if is_ref_attr_or_arn:
            if attribute == "Arn" and resource_state.get("QueueArn"):
                return resolve_refs_recursively(stack, resource_state.get("QueueArn"))
            return aws_stack.get_sqs_queue_url(resource_props.get("QueueName"))
    attribute_lower = first_char_to_lower(attribute)
    result = resource_state.get(attribute) or resource_state.get(attribute_lower)
    if result is None and isinstance(resource, dict):
        result = resource_props.get(attribute) or resource_props.get(attribute_lower)