Are the path integral formalism and the operator formalism inequivalent?
General comments to the question (v1):
Any textbook derivation of the correspondence between $$\tag{1} \text{Operator formalism}\qquad \longleftrightarrow \qquad \text{Path integral formalism}$$ is just a formal derivation, which discards contributions in the process, cf. e.g. this Phys.SE post.
Rather than claiming complete understanding and existence of the correspondence (1), it is probably more fair to say that we have a long list of theories (such as e.g. Yang-mills, Cherns-Simons, etc.), where both sides of the correspondence (1) have been worked out.
The correspondence (1) is mired with subtleties. Example: Consider a non-relativistic point particle on a curved target manifold $(M,g)$ with classical Hamiltonian $$\tag{2} H_{\rm cl} ~=~\frac{1}{2} p_i p_jg^{ij}(x), $$ which we use in the Hamiltonian action of the phase space path integral. Then one may show that the corresponding Hamiltonian operator is $$\tag{3}\hat{H}~=~ \frac{1}{2\sqrt[4]{g}} \hat{p}_i\sqrt{g}~ g^{ij} ~\hat{p}_j\frac{1}{\sqrt[4]{g}}+ \frac{\hbar^2R}{8} +{\cal O}(\hbar^3),$$ cf. Refs. 1 & 2. The first term in eq. (3) is the naive guess, cf. my Phys.SE answer here. The two-loop correction proportional to the scalar curvature $R$ is a surprise, which foretells that a full understanding of the correspondence (1) is going to be complicated.
References:
F. Bastianelli and P. van Nieuwenhuizen, Path Integrals and Anomalies in Curved Space, 2006.
B. DeWitt, Supermanifolds, Cambridge Univ. Press, 1992.
Short answer to your question: yes. There's no reason for the path integral and the operator formalism to be equivalent.
A simple example are all the non-Lagrangian theories. We know (some of) them through their operator algebra, but there's no corresponding Lagrangian. This might sound strange to most people, like myself not long ago, but it is not hard to understand. Lagrangian are only defined through their classical action (the limit $\hbar \to 0$,) which is nothing but the weak coupling limit (I've realized that many QFT books do not mention this fact.) If you have a theory that only lives at a strongly coupled fixed point, then you cannot construct a Lagrangian. As an example of this, just google "non Lagrangian theories." In my case, I get this article http://arxiv.org/abs/1505.05834 with a very nice introduction.
This is just to answer the title.
Regarding the main body, I think you are confused. First, you have to understand that the equivalence of the path integral and the operator formalism that you see in your QFT textbook is not by chance, it is by construction. So no, you are never going to spam a different Hilbert space by working with two formalism of the "same theory." Point (2) of Qmechanic is accurate in this sense. Since you know how to formulate the PI formalism for those theories, you are going to get the same answer.
Second, you argued that the LFH of (10) might be different. Truth is, it can't. The reason is simple: the propagator just move a state in space-time, it does not change it, it is by definition. Therefore, you can only get $1_{\mathcal H}$. If you ever get something else, then you are looking at the wrong object as the propagator.
Finally, you must always be careful when talking about the Path Integral of anything: it is full of subtle points. Actually, as of today, it is not know what is QFT, so you should also be careful when talking about the Operator Formalism of anything!