I think that the answer to your question is no, a transaction which makes every block it is included in invalid does not exist.
An important core property of a cryptographic hash function is that there is no discernable relationship between the input data, and output data. If such a relationship existed, then miners could 'cheat' by only creating blocks (input data) that would lead to a valid hash (output data) in the preferred range (lower than the difficulty target). In reality, we see that changing the input data slightly will drastically alter the output data, in a deterministic but unpredictable way.
And so I don't believe a transaction could exist, which would invalidate every block which it was attempted to be included in. Even if you create a block with just a coinbase transaction, and exhaust the nonce and extranonce space without finding a valid solution, the timestamp will still slowly increment, and each time it does you can search the entire nonce/extranonce space again. Probabilistically, since we expect an even, and random output value distribution from the hash function, after enough time, some combination of your transaction / the timestamp / the nonce and extranonce should lead to a valid block.
There is also lots of other data that can be included in a block, and altering any piece of it will affect the block hash as well. Murch's answer here tells us that lately there are ~2000 transactions in a block on average. For simplicity, lets assume that all transactions in the block are independent, so they could be included in any order. The number of permutations for the transaction ordering is thus
2000!, which is a huge number (~3.316 E +5735). For each of these combinations, the miner could exhaust the nonce space, the extranonce space, and then increment the timestamp and do it all again. This makes our number of possible arrangements even more enormous! Given enough time, and computational power, it becomes likely that a valid block hash will be found.
So, just from a probabilistic argument, I don't believe it is possible for the inclusion of some transaction to magically only create invalid blocks. Perhaps someone more well versed in the nitty-gritty details can offer a more thorough proof of this, but I think just the numbers alone, combined with the security properties of a cryptographic hash function, are reasonable enough proof.