Is One Time Pad considered Chosen-Plaintext Attack Secure? If we're considering Chosen-Plaintext Attack setting, then the adversary has access to the Encryption Oracle right, and we know that OTP is only considered secure if we use the key only once. How would we ensure that the adversary uses the key only once?

Or would it not matter how the adversary uses the encryption oracle since in the end they will be presented with a ciphertext corresponding to a plaintext not queried and the adversary would not be able to decrypt it because they don't know the key. Making it CPA-Secure. Is this second train of thought the correct way to think about OTP in CPA setting?

Thank you!
Note: In this answer, I stick to a definition of the One Time Pad where the random pad is used only One Time; at least, I've the name of it as support! Otherwise, it is well known that the OTP encryption scheme consisting of XOR with a repeated key is insecure by even the weakest standard (unknown plaintext with redundancy).

Late addition: further, I stick to the definition of IND-CPA given for ciphers in standard literature and quoted below. See this other answer for a different definition, which applies to the OTP, and is met.

INDistinguishability under Chosen Plaintext Attack applies to encryption schemes; there's no way the OTP can be made to match the definition of an encryption scheme, thus the formal definition of IND-CPA does not apply to the OTP.

Specifically, the OTP does not fit definition 3.7 in Jonathan Katz and Yehuda Lindell's Introduction to Modern Cryptography, first edition (CRC Press, 2008):


  A private-key encryption scheme is a tuple of probabilistic
  polynomial-time algorithms ($\text{Gen}$, $\text{Enc}$, $\text{Dec}$) such that: (..)
  
  
  The key-generation algorithm $\text{Gen}$ takes as input the security parameter $1^n$ and outputs a key $k$; (..)
  The encryption algorithm $\text{Enc}$ takes as input a key $k$ and a plaintext message $m\in\{0,1\}^*$, and outputs a ciphertext $c$. (..)
  The decryption algorithm $\text{Dec}$ takes as input a key $k$ and a ciphertext $c$, and outputs a message $m$. (..)
  
  
  It is required that for every $n$, every key $k$ output by $\text{Gen}(1^n)$, and every $m\in\{0,1\}^*$, it holds that $\text{Dec}_k(\text{Enc}_k(m))=m$.


where $\{0,1\}^*$ is the set of all (finite-length) binary strings.

There's no way the OTP can fit this framework:


If we try to assimilate the random pad and the key, we hit the problem that in the above definition the same $k$ can be used for several messages (and further that some messages can be larger than the key, but that's secondary); the OTP does not allow that (due to the One Time requirement of OTP).
If we try to fit the random pad into $\text{Enc}$, and still keep the ciphertext as it is in the OTP (including, of the same size as the message), then we have no way to build a working $\text{Dec}$.


Note: some practical encryption schemes use message spaces deviating from $\{0,1\}^*$, like the set of all bytestrings up to some size, or even a fixed-size bytestring, perhaps even shorter than the key (which is common in block ciphers); and it is accepted to twist the definition of an encryption scheme to allow that. However it is an essential point of any encryption scheme that the total size of plaintext enciphered with a given key is allowed to increase significantly above the size of the key; again, the OTP does not allow that.