If $N$ is embedded in $M$, then there exists a superstructure $N \subset N' \cong M$

Question

I have a language $L$ and two $L$-structures $N,M$ and a function $\varphi$ such that $$\varphi:N \hookrightarrow M$$ is an embedding. I want to prove that there exists $N'$ superstructure of $N$ that is isomorphic to $M$.

The sketch of the proof is the following:

1. I define a set $N'$ by taking $M$ and substituing $\varphi(N)$ with $N$, that is: $$N' = N \cup (M \setminus \varphi(N))$$ and also define a bijection $\phi: N' \rightarrow M$ s.t. $$\phi(x) = \begin{cases} f(x) & \text{if $x \in N$} \\ x & \text{otherwise} \\ \end{cases}$$
2. I endow $N'$ with an $L$-structure through $\phi$. In this way $\phi$ is going to be an isomorphism.
3. Prove that $N$ is indeed a substructure of $N'$, by showing for example that the inclusion map $i: N \rightarrow N'$ is an embedding, which is, as $$i = \phi^{-1} \circ f$$

Now: is this correct? Is there a quicker way to do this? Thanks

EDIT:

The answer of Noah Schweber pointed out that there is a serious problem when $(M\setminus \varphi(N)) \cap N \neq \emptyset$. Indeed in this case the function $\phi$ is no longer surjective. If I find a structure $M'$ such that $M' \cap N = \emptyset$ and $M \cong M'$, then I'm done.

This is done by finding a set $M'$ such that $M' \asymp M$ (with $\psi:M\rightarrow M'$ bijection) and $M' \cap N = \emptyset$, since then I can endow $M'$ with an $L$-structure through $\psi$, making $\psi$ an isomorphism.

I can always find such set trough replacement, by defining $\psi: M \rightarrow V$ as: $$\psi(x) = \{N,x\}$$ in this way $M'=\psi(M)$ is a set equivalent to $M$ and, by foundation, $M' \cap N = \emptyset$.

Answer 1

This is the right idea, but there's an odd subtlety here: what if $N\cap (M\setminus\varphi(N))\not=\emptyset$? That is, what if $M$ uses a thing which happens to be an element of $N$, but in a different way (= not in the image of $\varphi$)? For example, working in the empty language (so structures are just sets), consider $N=\{1\}$, $M=\{1,2\}$, and $\varphi: N\rightarrow M: 1\mapsto 2$.

In this situation the idea you outline above won't work; you need to massage things a bit to avoid this kind of overlap. But the main idea you have of just "copying" the rest of $M$ is definitely right in spirit.

(HINT: after checking where exactly your argument breaks down due to the above issue, show that it does work if we also assume $M\cap N=\emptyset$. This reduces the problem to finding an $M'\cong M$ with $M'\cap N=\emptyset$ ...)