Can AES-CTR add values without decryption?

I recently have studied AES cipher. On CTR mode, is it possible to add, subtract, or XOR plaintext values on encrypted data?

For example: we have an encryption of $4$ and $6$ under the same AES key, respectively denoted by $AES(4)$ and $AES(6)$.

Without decryption, can we achieve the following?

$$AES(4)⊕AES(6)=AES(4+6)$$

Where "$\oplus$" is any operation we can do over AES.

nor define operator $\boxplus$ such that $\operatorname{AES-CTR}(4)\boxplus\operatorname{AES-CTR}(6)=\operatorname{AES-CTR}(4+6)$ that works for unknown key and plaintexts. Informally: if that worked, it would be a weakness of the encryption scheme, which would be leaking information about plaintexts from the ciphertexts.
Notice that the above notation is misleading, as the result of AES-CTR depends on 3 parameters: Key, IV and plaintext, with IV normally random (making the argument above non-rigorous). But still, even if we assume fixed IV, what's asked can't be done: it is asking an operator $\boxplus$ such that $\forall s,\forall x,\forall y, (s\oplus x)\boxplus(s\oplus y)=s\oplus(x+y)$ where $\oplus$ is XOR, and $s$ is whatever the IV encrypts to under the key considered; which is impossible.
• @Gilles: The argument can be made rigorous for unique-counter. The one(s) for $\operatorname{AES-CTR}(x+y)$ or $\operatorname{AES-CTR}(x\oplus y)$ can't be random, and the only choice that could work reliably are this/those used for one of $\operatorname{AES-CTR}(x)$ or $\operatorname{AES-CTR}(y)$ (in the random cipher model, almost nothing can be said about how other counters would encrypt); and then almost nothing is known about whichever of $x$ or $y$ uses the other counters, which makes a suitable $\boxplus$ hopeless. – fgrieu May 14 '17 at 18:17