Cracking RSA ciphertext without a public key
It is asked if in RSA, without the public key, 


  it is likely much more difficult (if not almost impossible) to find the correct prime numbers, the value $e$, and the value $d$.


As first noted by Tylo in comment, if the RSA key was chosen competently, it will be impossible to find the prime numbers and $d$. Argument: if there was a method making that possible, then that method would also break regular RSA (with the public key available to the attacker); and that's a well-studied and unsolved problem, including if RSA is otherwise used incompetently (like, without padding as in textbook RSA).

Wether it will be possible to find $e$ depends on how $e$ was chosen, on the RSA padding, and on if sufficiently many known plaintext-ciphertext pairs are available:


If $e$ was chosen at random in a large enough set (e.g. as a random odd integer with $2&lt;e&lt;(p-1)(q-1)$ and $\gcd(e,p-1)=1=\gcd(e,q-1)$ ), then it can't be found (argument: that $e$ is also the private exponent in an RSA system with the public exponent $d$, and we just said the adversary can't find the private exponent).
If $e$ is small, it can be guessed (much more if it is among the usual values $e=2^{(2^k)}+1$ with $0\le k\le4$, and $e=37$ ).
If additionally there is no random padding (as in textbook RSA), and a number of known plaintext-ciphertext pairs $(m_i,c_i)$ is known, then a guess of $e$ can be verified, and $N=p\;q$ found, by computing the GCD of enough ${m_i}^e-c_i$ (where the exponentiation is not modular). Because $N$ divides ${m_i}^e-c_i$, for enough plaintext-ciphertext pairs, this GCD will be $1$ for an incorrect guess of $e$, and $N$ for a correct guess.