I guess $G/N\cong G'/N'$ doesn't hold in this case, but I cannot prove that.

Question

I guess the following proposition is false.
But I cannot prove that the following proposition is false.

Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $G'$ be a group.
Let $N'$ be a normal subgroup of $G'$.
Suppose that $G\cong G'$.
Suppose that $N\cong N'$.
Then, $G/N\cong G'/N'$ holds.

Intuitively, I think the following proposition is true.

Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $G'$ be a group.
Let $N'$ be a normal subgroup of $G'$.
Suppose that $G\cong G'$ and $f$ is an isomorphism from $G$ to $G'$.
Suppose that $N\cong N'$ and $f|_N$ is an isomorphism from $N$ to $N'$.
Then, $G/N\cong G'/N'$ holds.

Answer 1

Consider the case $G = G' = \mathbb{Z}$. Then $N = \langle 2 \rangle$ and $N' = \langle 3 \rangle$ are isomorphic since they're both infinite cyclic, but their quotients are not as they are $\mathbb{Z}_2$ and $\mathbb{Z}_3$ respectively.

Your refined version is correct. Define a map $f': G \to G'/N'$ by $f'(x)=f(x)N'$. The kernel will be $N$ by your assumptions, and since this is surjective you're done.