您需要在计算总计的子查询中联接order和order_products。并且这需要按客户ID分组。然后，您将其与主查询中的客户表联接。任何聚合函数都不能在主查询中，因为这样它们将乘以您要联接的地址数。
您需要一个单独的子查询，该子查询只从order聚合，而不从order_products聚合，这样这些聚合就不会乘以乘积。
您不需要在主查询中进行任何分组，因为主查询中现在没有聚合函数。您将获得每个客户地址的单独结果行--订单信息相同，但地址不同。

SELECT 
    c.customer_id, 
    c.customer_lastname, 
    c.customer_firstname, 
    o2.sold_products_count,
    o2.returned_products_count,
    ROUND(100 * returned_products_count / sold_products_count) AS returned_products_percent,
    ROUND(100 * SUM(IF(op.product_id > 0, IF(op.product_quantity < 0, -1 * op.product_quantity, 0), 0)) 
    / SUM(IF(op.product_id > 0, IF(op.product_quantity > 0, op.product_quantity, 0), 0))) as returned_products_percent, 
    o1.orders_total,
    o1.orders_count,
    ci.customer_street_address, 
    ci.customer_phone,
    o1.first_order_date,
    o1.last_order_date,
    o1.days_since_last_order
FROM customer AS c
LEFT JOIN customer_address AS ci USING (customer_id) 
LEFT JOIN (
     SELECT
        o.customer_id,
        MIN(o.order_datetime) as first_order_date, 
        MAX(o.order_datetime) as last_order_date, 
        DATEDIFF(NOW(), MAX(o.order_datetime)) as days_since_last_order, 
        COUNT(*) as orders_count, 
        SUM(o.order_total)  as orders_total
    FROM `order` AS o
    GROUP BY o.customer_id) AS o1 ON o1.customer_id = c.customer_id
LEFT JOIN (
    SELECT
        o.customer_id,
        SUM(IF(op.product_id > 0, IF(op.product_quantity > 0, op.product_quantity, 0), 0))  AS sold_products_count, 
        SUM(IF(op.product_id > 0, IF(op.product_quantity < 0, -1 * op.product_quantity, 0), 0)) AS returned_products_count
    FROM `order` as o
    JOIN order_products AS op ON o.order_id = op.order_id
    GROUP BY o.customer_id) as o2 ON o2.customer_id = c.customer_id
GROUP BY c.customer_id 
ORDER BY c.customer_id ASC
limit 100