The concept of an immersed manifold often leads to confusion and I do not see a real benefit to use it.
Let us consider a more general situation: Given a smooth manifold $M$, a set $S$ and a bijection $f : N \to S$. Then it is clear that
there exists a unique topology $\tau$ on $S$ making $f : N \to (S,\tau)$ a homeomorphism.
there exists a unique smooth structure $\mathfrak D$ on $(S,\tau)$ making $f : N \to (S,\tau,\mathfrak D)$ a diffeomorphism.
Lee uses this with $S = F(N)$. His definition of open sets and of charts is nothing else than the explicit construction of $\tau$ and $\mathfrak D$.
The problem is that $\tau$ in general is not the subspace topology inherited from $M$. As an example take the injective smooth immersion $\beta : N = (-\pi,\pi) \to \mathbb R^2$ decribed here. Its image $S$ is a "figure eight" which is compact and thus cannot be homeomorphic to $N$. Moreover $S$ is not a manifold. Thus, unfortunately, your approach does not work.
You see that the "immersed submanifold structure" of $S$ is a fairly artificial thing which has nothing to do with the subspace $S$ of $\mathbb R^2$.
Also have a look at Restricting the codomain of a smooth map to submanifold is smooth.
Update:
Let us consider the bijective function $\bar F : N \stackrel{F}{\to} F(N)$. We endow $F(N)$ with the unique topology and smooth structure making $\bar F$ a diffeomorphism. If $\iota : F(N) \hookrightarrow M$ denotes inclusion, then clearly $\iota \circ \bar F = F$. Since $\bar F$ is a diffeomorphism, we get $\iota = F \circ \bar F^{-1}$. This shows that $\iota$ is an injective immersion because $F$ is one.
I believe I have found a uniqueness result for the smooth structure in the case of an embedded submanifold with boundary.
The following theorem is an analogue of Theorem 5.31 (Uniqueness of Smooth Structures on Submanifolds) for embedded submanifolds with boundary. However it only claims that the smooth structure is unique once we assume the subspace topology, unlike Theorem 5.31 which also shows that the topology is unique even if we only assume the submanifold is immersed.
In fact, the topology is not unique for submanifolds with boundary, because for example the unit circle is an embedded submanifold with boundary in $\mathbb{R}^2$ (with empty boundary) but it is also an immersed submanifold with boundary with the topology given by the injective smooth immersion $\gamma: [0, 1) \to \mathbb{R}^2$ such that $\gamma(t) = (\cos 2\pi t, \sin 2\pi t)$. This makes it a submanifold with boundary where the boundary is $\{(1, 0)\}$, and there are open subsets containing $(1, 0)$ that are not open in the subspace topology, namely the images $\gamma([0, a))$ for $0 \lt a \lt 1$.
Since Theorem 5.51 is only stated for $M$ a smooth manifold without boundary, we cannot apply it in the case where $M$ itself has nonempty boundary.
Theorem
Suppose $M$ is a smooth manifold without boundary, and $S$ is an embedded submanifold with boundary. The smooth structure on $S$ described in Theorem 5.51 is the only smooth structure with respect to which $S$ is an embedded submanifold with boundary.
Proof
Most of the proof is copied verbatim from the proof of Theorem 5.31, with minor modifications to use the theorems for restricting the codomains of smooth maps between manifolds with boundary, and to exclude the possibility of a different topology. But we cannot use the Global Rank Theorem to show that $\widetilde{\iota}$ is a diffeomorphism. Instead, we show directly that it has a smooth inverse.
Suppose $S \subseteq M$ is an embedded $k$-dimensional submanifold with boundary. Theorem 5.51 shows that it satisfies the local $k$-slice condition for submanifolds with boundary, so it is an embedded submanifold with boundary with the subspace topology and the smooth structure of Theorem 5.51. Suppose there was some other smooth structure making it into an embedded submanifold with boundary. Let $\widetilde{S}$ denote the same set $S$, considered as a smooth manifold with boundary with the non-standard smooth structure, and let $\widetilde{\iota}: \widetilde{S} \hookrightarrow M$ denote the inclusion map, which by assumption is a smooth embedding. Because $\widetilde{\iota}(\widetilde{S}) = S$, Theorem 5.53 (b) implies that $\widetilde{\iota}$ is also smooth when considered as a map from $\widetilde{S}$ to $S$.
Since $\widetilde{S}$ is also an embedded submanifold with boundary, it also has the subspace topology. Therefore, $\widetilde{\iota}: \widetilde{S} \to S$ is a homeomorphism, so $\widetilde{\iota}^{-1}: S \to \widetilde{S}$ is continuous. It is the map obtained by restricting the codomain of the smooth map $\iota: S \hookrightarrow M$ to $\widetilde{S}$, which is an embedded submanifold with boundary in $M$. Therefore, Theorem 5.53 (b) also implies that $\widetilde{\iota}^{-1}: S \to \widetilde{S}$ is smooth.
These results show that $\widetilde{\iota}: \widetilde{S} \to S$ is a diffeomorphism and so $\widetilde{S}$ has the same smooth structure as $S$.
Best Answer
No, your proof is not right because you didn't link it to the manifold structure of the product, all you showed is there is a smooth structure on the graph for which your $\Phi$ is a diffeomorphism. This is actually very obvious: given a type of mathematical object $M$ (group/ring/field/vector space/inner-product space/Banach space/smooth manifold etc), any set $S$, and a bijection $\Phi:M\to S$, we can always 'transport' the structure from $M$ to $S$ such that $\Phi$ becomes an isomorphism in the appropriate sense.
You can actually generalize the statement. If $f:M\to N$ is a smooth map between smooth manifolds, then its graph, $\Gamma(f)$, is an embedded submanifold of $M\times N$ (equipped with the product manifold structure). You can also weaken the hypothesis to $C^r$ manifolds and $C^r$ maps, and the proof is oblivious to whether you consider finite-dimensional manifolds or Banach manifolds.
To prove this, take any point $(p,f(p))\in \Gamma(f)$. Fix a chart $(U,\alpha)$ in $M$ around the point $p$ and a chart $(V,\beta)$ around the point $f(p)$ in $N$, and also assume that $f$ maps $U$ into $V$ (we can always arrange for this since $f$ is continuous so we can replace $U$ by the smaller open set $U\cap f^{-1}(V)$ if necessary). This gives rise to a 'product chart' $(U\times V,\alpha\times \beta)$, in the obvious way $\alpha\times \beta:U\times V\to \alpha[U]\times\beta[V]$, $(x,y)\mapsto (\alpha(x),\beta(y))$. Now, let us construct a new chart on the product manifold, $(U\times V, \psi_f)$, where the map $\psi_f$ is defined as \begin{align} \psi_f(x,y):=(\alpha(x),\beta(y)-\beta(f(x))) \end{align} I leave it to you to justify why $(U\times V,\psi_f)$ actually belongs to the maximal atlas on the product (i.e you must show why it is compatible with all other product charts). This is easy to do by-hand because inverting things is very trivial here. Now, the chart $(U\times V,\psi_f)$, when restricted to $\Gamma(f)$ is such that is maps $\Gamma(f)\cap (U\times V)$ onto $\psi_f[U\times V]\cap \alpha[U]\times \{0\}$, i.e $\psi_f$ flattens out the graph in the obvious way, by subtracting the height of the graph at every point. Thus, I've shown that for each point in the graph, there is a chart belonging to the maximal atlas of the product $M\times N$, such that this chart satisfies the 'slice-condition' in (one of the equivalent) definition of an embedded submanifold.
Here, I gave the argument explicitly (leaving minor details for you to fill in) because it is possible to do this without any heavy lifting. There is a more general theorem, as stated and proved here, relying on the inverse/implicit function theorem. Note that in the link, I stated it for finite-dimensional manifolds, but it generalizes easily to the Banach setting (perhaps one needs to add some conditions regarding existence of closed complementary subspaces), because the inverse function theorem holds in Banach spaces (because one of the ingredients in proving the IFT is Banach's fixed point theorem which holds in complete metric spaces... e.g Banach spaces:) If you want to apply the theorem quoted, then you just have to verify that $\Phi:M\to M\times N$, $x\mapsto (x,f(x))$ is an immersion, and it is a homeomorphism onto its image, which is the graph $\Gamma(f)$. I leave these two as exercises:)