First let me point out a small typo : it should be functors over $\mathcal C$ which send cocartesian edges to cocartesian edges (this doesn't matter for left fibrations, but for cocartesian fibrations it does).
For your main point, you are right that they are different quasi-categories: the simplicial sets are not isomorphic; but they're the same $\infty$-category, that is, the obvious forgetful functor from one to the other (so the one that uses the trivial $1$-simplex) is a categorical equivalence.
The fact that it is essentially surjective is obvious (the objects are the same), so the point is really about mapping spaces. But this is really a statement about pullbacks, namely the following:
Consider three quasi-categories $C,D,E$ with functors $p: C\to E, q:D \to E$. Suppose $q: D\to E$ is a cocartesian fibration. In this situation, the quasi-category $Fun_E(C,D)$ defined as a simplicial set by the (ordinary) pullback $Fun(C,D)\times_{Fun(C,E)}\{p\}$ is equivalent to the homotopy pullback, $Fun(C,D)\times_{Fun(C,E)}(Fun(C,E)^\simeq)_{/p}$
Note that the first one is the category of those functors $f :C\to D$ that are strictly over $E$, that is, with $q\circ f = p$, whereas $0$-simplices in the latter are a pair $(f, \sigma)$ where $f : C\to D$ is a functor and $\sigma: q\circ f \to p$ is an equivalence, and you can check that in fact it is what you're thinking of.
The reason for this equivalence is that $Fun(C,D)\to Fun(C,E)$ is also a cocartesian fibration, so in particular a fibration in the Joyal model structure, and so pull-back along it preserves weak equivalences. It then suffices to observe that the inclusion $\{p\}\to (Fun(C,E)^\simeq)_{/p}$ is an equivalence, because $Fun(C,E)^\simeq$ is a Kan complex.
The idea is that because $D\to E$ is a cocartesian fibration, you have "enough room" to "strictify" any equivalence $q\circ f\simeq p$ to an actual equality $q\circ f = p$ up to changing $f$.
(note that the latter, the one with $q\circ f\simeq p$, is the one that you would want to have as an $\infty$-category, because $=$ is too strict a notion. It just happens that in this case you can actually strictify)
Best Answer
The answer is yes, but $\mathrm{LFib}(\mathcal C)$ is also the full subcategory of $(\mathrm{Cat}_\infty)_{/\mathcal C}$, it just so happens that you can prove that any morphism between such is a left fibration (this does not remain true in the case of cocartesian fibrations, though).
Here is a proof: Let $f:\mathcal{D\to E}$ be a morphism of left fibrations over $\mathcal C$, which I'll denote by $p,q$ respectively. By the dual of HTT.2.4.1.3.(3), if an edge $\alpha$ in $\mathcal D$ is such that $f(\alpha)$ is $q$-coCartesian, then $\alpha$ is $f$-coCartesian if and only if it is $p$-coCartesian.
But we are in left fibrations, so all edges are $p$-coCartesian. In particular, $\alpha$ is $f$-coCartesian if and only if $f(\alpha)$ is $q$-coCartesian, the latter being a void condition: every edge is $f$-coCartesian.
So we have seen that any edge is $f$-coCartesian, and therefore we just need a sufficient supply of (arbitrary) edges along $f$: let $\alpha: e_0\to e_1$ be an edge in $\mathcal E$, with a lift $d_0$ of $e_0$ along $f$, i.e. $f(d_0) = e_0$. We then push things down: $q(\alpha): q(e_0) \to q(e_1)$ is an edge in $\mathcal C$, with source $q(e_0) = p(d_0)$, so it has a lift $\tilde \alpha: d_0\to d_1$. Now $f(\tilde\alpha): f(d_0)\to f(d_1)$ is some map lifting $q(\alpha)$, and it is $q$-coCartesian, just like $\alpha$ (and any map in $\mathcal E$), so there is an equivalence $e_1\simeq f(d_1)$ with appropriate $2$-simplices showing that have found a lift of $\alpha$.