This is a great question and something that confused me while studying superconductivity, where we often think of bound states of two electrons as a boson. I think the other answer and comments by mmesser314 captures the gist of the idea, but let me elaborate.
Your flow of logic in points 1,2,3 are perfect. The idea works when your assumption in point 1 is not valid, i. e. the two fermions are not in a product state and are instead entangled.
To explain this, I will introduce the occupation number formalism and the language of second quantization (but very simplified, I will sacrifice precision to get the point across easily).
Occupation number formalism
Instead of thinking in terms of particles and their wavefunctions, let us think of quantum states in terms of the "orbitals" and whether they are occupied or not. This is useful and natural when we are dealing with identical particles. For simplicity, let us think of an electronic system with $n$ orbitals, where each orbital can have both spin-$\uparrow$ and spin-$\downarrow$ electrons. So the empty quantum state with no electrons would be written as
$$
\vert \rangle
$$
and a state with one spin-$\downarrow$ electron in site 3 and one spin-$\uparrow$ electron in site 5 would be represented as
$$
\vert \downarrow_3 \uparrow_5 \rangle
$$
and so on. This is the occupation number representation of quantum states with many particles. We say two electrons are entangled if the state is in some superposition
$$
\vert \downarrow_3 \uparrow_5 \rangle + \vert \downarrow_4 \uparrow_7 \rangle
$$
since if we measure a spin-down electron at orbital 4, we gain information about the location and spin of the other electron.
Creation and annihilation operators
A useful object in this language (sometimes called second quantization) are operations that act on quantum states to create ($c$) or remove ($a$) particles. Let us define
$$
c_{\downarrow_7}\vert \downarrow_3 \uparrow_5 \rangle = \vert \downarrow_3 \uparrow_5 \downarrow_7 \rangle \qquad a_{\downarrow_3}\vert \downarrow_3 \uparrow_5 \rangle = \vert \uparrow_5 \rangle
$$
Because of Pauli's exclusion principle for fermions, we want them to satisfy
$$
c_i c_i \vert \psi \rangle = 0 \qquad a_i a_i \vert \psi \rangle = 0
$$
for any state $\vert \psi \rangle$, as we cannot put two electrons on the same single-particle state $i$ (which now includes both orbital and spin), which also means that we cannot remove two electrons from a single-particle state. The right-hand-side being zero (not to be confused with the empty state $\vert \rangle$) tells us that this is an invalid operation.
If I put your intuitive idea into this language, this is how you would say it. I want to think of a state with two electrons, say $\vert \uparrow_1 \downarrow_1\rangle$ as a boson. In other words, we want to define the boson creation operator
$$
c_b = c_{\uparrow_1} c_{\downarrow_1}
$$
But this does not work! If I try to put two bosons in this state, I end up getting $0$ because of Pauli's exclusion principle.
$$
c_b c_b \vert \rangle = c_{\uparrow_1} c_{\downarrow_1} c_{\uparrow_1} c_{\downarrow_1} \vert \rangle = c_{\uparrow_1} c_{\downarrow_1} \vert \uparrow_1 \downarrow_1\rangle = 0
$$
So that is not allowed, and we cannot think of this two-particle state as a boson.
The trick is to put the two electrons in a large superposition. Let us define the two particle state to be
$$
\vert \uparrow_1 \downarrow_1 \rangle + \vert \uparrow_2 \downarrow_2 \rangle + \vert \uparrow_3 \downarrow_3 \rangle + \dots +\vert \uparrow_n \downarrow_n \rangle,
$$
Equivalently, let us define the boson creation operator as
$$
c_b = c_{\uparrow_1} c_{\downarrow_1} + c_{\uparrow_2} c_{\downarrow_2} + c_{\uparrow_3} c_{\downarrow_3} + \dots + c_{\uparrow_n} c_{\downarrow_n}
$$
Now we can think of this as a boson (approximately), because
\begin{align}
c_b c_b \vert \rangle &= c_b (\vert \uparrow_1 \downarrow_1 \rangle + \vert \uparrow_2 \downarrow_2 \rangle + \vert \uparrow_3 \downarrow_3 \rangle + \dots +\vert \uparrow_n \downarrow_n \rangle ) \\
&= \sum_{i \neq j} \vert \uparrow_{i} \downarrow_{i} \uparrow_{j} \downarrow_{j} \rangle
\end{align}
In fact, you can convince yourself that you can add up to $n$ bosons in this fashion without violating Pauli's exclusion principle! So, as long as your number of "bosons" is much less than $n$, it is okay to think of this two particle bound superposition state as approximately a boson!
Summary: The misunderstanding in your argument is that, when the constituent fermions are entangled, the statement "if A and B have the same quantum state, then by 1 the constituent particles must have the same quantum state" is meaningless. You cannot think of the constituent particles being in a definite quantum state if they are entangled.
