Can elliptic curve groups be used for commutative encryption?

In trying to implement mental poker, can all players agree on a standard set of 52 points on the curve corresponding to each card, and then to "encrypt" a card you just multiply it by a scalar which is your encryption key? (and to decrypt, multiply by the scalar's inverse modulo the group size)

This works, as long as the 52 points are chosen such that their discrete logs w.r.t. each other are unknowable.

E.g. if $5♠$ is point $P$, and $6♦$ is point $Q$, and your encryption scalar is $x$, then you could encrypt $5♠$ using $P' = xP$. Now, if you know $y=P/Q$, then you could pretend you had encrypted $6♦$ instead of $5♠$ by declaring your encryption key as $z=xy$, because $z^{-1}P'=(xy)^{-1}P'=(xP/Q)^{-1}P'=(xP/Q)^{-1}xP=(Q/xP)xP=Q$

I'd recommend you use a hash-to-curve method to produce the point $P$ by hashing the string $5♠$. The discrete logs of hash-to-curve points with respect to other curve points are unknowable.

In contrast, it would be catastrophic if you'd chosen points such that each point was a base point $G$ added to itself between 1 and 52 times, depending on the card.

Note that you would need to use a different encryption key every time you encrypt a card, to ensure that the same card encrypted twice (during different rounds) does not produce the same encrypted point each time.