Yes, it is sufficient. Your question actually touches on a known, similar issue. A proper class is not an object in ZF, so in ZF we can't directly refer to it. Hence, again under the assumption we work in ZF, $V$ is not an object, $j$ is not an object, and if $M$ is a proper class, $M$ is also not an object. Despite all this, mathematicians working in ZF talk about these things indirectly.
The trick is, one can assume, with an exception to the case where $j \colon V \to V$, that classes $M$ and $j$ are definable from parameters by $\Sigma_2$ formulas. Moreover, the assertion that $j$ is an elementary embedding is equivalent to the assertion that for all formulas $\varphi(x)$, all ordinals $\alpha$, and all sets $a \in V_\alpha$ $$V_\alpha \models \varphi[a]$$ if and only if $j(V_\alpha) \models \varphi[j(a)]$.
So, getting back to your original question, the answer is yes. One can talk about an elementary embedding $j \colon V \to M$ without accepting the existence of $V$. The method is the same as the one outlined above.
Another way of looking at your problem is to think of a less complicated, similar scenario. In classical potentialism, one refutes the existence of $\mathbb{N}$ but accepts all its finite initial segments. Here, the problem would be whether one can talk about functions $f \colon \mathbb{N} \to A$, where $A$ is some finite set of natural numbers. In particular, let us call $n$ a pigeonhole number whenever there is an $m<n$ such that $f(m) = f(n)$.
The question: For every finite set of natural numbers $A$, does every function $f \colon \mathbb{N} \to A$ admit a pigeonhole number? While $f$ and $\mathbb{N}$ don't exist for a classical potentialist, he can still answer this question! He can't do it explicitly, but he can do it implicitly using larger and larger finite initial segments of $\mathbb{N}$.