Aunque no estoy seguro de qué significa "problemático" en este contexto, aquí está la consulta reescrita como un simple LEFT JOIN
con una subconsulta solo para obtener la clasificación correcta al final (el ORDER BY
debe hacerse antes de la clasificación);
SELECT user_id, score, @rank := @rank + 1 AS rank FROM
(
SELECT u.user_id, u.score
FROM user_score u
LEFT JOIN user_score u2
ON u.user_id=u2.user_id
AND u.`timestamp` < u2.`timestamp`
WHERE u2.`timestamp` IS NULL
ORDER BY u.score DESC
) zz, (SELECT @rank := 0) z;
EDITAR:para tener en cuenta group_id, deberá ampliar un poco la consulta;
SELECT user_id, score, @rank := @rank + 1 AS rank FROM
(
SELECT u.user_id, u.score
FROM user_score u
LEFT JOIN user_score u2
ON u.user_id=u2.user_id
AND u.group_id = u2.group_id -- u and u2 have the same group
AND u.`timestamp` < u2.`timestamp`
WHERE u2.`timestamp` IS NULL
AND u.group_id = 1 -- ...and that group is group 1
ORDER BY u.score DESC
) zz, (SELECT @rank := 0) z;