Your notation is OK up to and including step 2, except for the range of summation. You need
$$
|s\rangle=\frac{1}{\sqrt{2^{n+m}}}\sum_{x=0}^{2^n-1}\sum_{y=0}^{2^m-1}|x\rangle|y\rangle
$$
Now the problem is how to write down the effect of the oracle, and you cannot just write down the output qubit. I think you probably know this from the title of the question, because entanglement will appear that you're not describing. So, you have an oracle that acts as
$$
|x\rangle|y\rangle|0\rangle\xrightarrow{\text{oracle}}|x\rangle|y\rangle|F(x,y)\rangle.
$$
Hence, if the input is some superposition state such as $|s\rangle$, we have
$$
|s\rangle|0\rangle\xrightarrow{\text{oracle}}|\Psi\rangle=\frac{1}{\sqrt{2^{n+m}}}\sum_{x=0}^{2^n-1}\sum_{y=0}^{2^m-1}|x\rangle|y\rangle|F(x,y)\rangle.
$$
You absolutely cannot describe (except in very special cases of $F$) the last qubit in the form $\alpha|0\rangle+\beta|1\rangle$ because it is entangled with the other registers.
The question seems to be evolving into
If I'm not measuring $x$ or $y$, why can't the state of the extra qubit be written in the form $\alpha|0\rangle+\beta|1\rangle$?
There are several ways that this might be answered. Normally, I'd take the partial trace and calculate the reduced density matrix, but I infer from comments that the OP doesn't know this technique. Thus, let us try another route.
Let us assume that the extra qubit can be written in the form $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$. This means that we could define a measurement
$$
P_\psi=|\psi\rangle\langle\psi|\qquad P_{\perp}=|\psi^\perp\rangle\langle\psi^\perp|
$$
where $|\psi^\perp\rangle=\beta^{\star}|0\rangle-\alpha^\star|1\rangle$ is orthogonal to $|\psi\rangle$. If we can guarantee that the extra qubit is in that state, then we are guaranteed to get the measurement result $P_{\psi}$. In other words,
$$
\langle\Psi|\mathbf{I}\otimes\mathbf{I}\otimes P_{\psi}|\Psi\rangle=1.
$$
(The identity operations are how we say that we're not measuring the $x$ and $y$ systems.) I claim that there are no satisfying $\alpha,\beta$ where $|\alpha|^2+|\beta|^2=1$, unless $F(x,y)$ is a constant function.
So, we start to evaluate
\begin{align*}
\langle\Psi|\mathbf{I}\otimes\mathbf{I}\otimes P_{\psi}|\Psi\rangle&=\frac{1}{2^{n+m}}\sum_x\sum_y\langle F(x,y)|P_{\psi}|F(x,y)\rangle \\
&=\frac{1}{2^{n+m}}\left(\sum_{x,y:F(x,y)=0}|\alpha|^2+\sum_{x,y:F(x,y)=1}|\beta|^2\right) \\
&= \frac{1}{2^{n+m}}(M|\alpha|^2+(2^{n+m}-M)(1-|\alpha|^2))
\end{align*}
Where $M$ is the number of values such that $F(x,y)=0$. Setting this equal to 1, we can rearrange for $|\alpha|^2$:
$$
|\alpha|^2=\frac{M}{2M-2^{n+m}}
$$
For $|\alpha|^2$ to be a valid value, it must be $0\leq|\alpha|^2\leq 1$. One has to be careful in the analysis here. If we assume that $2M>2^{n+m}$, then the denominator is positive, and $|\alpha|^2\leq 1$ implies
$$
M\geq 2^{n+m}.
$$
This only happens if $M=2^{n+m}$, in other words, $F(x,y)=0$ for all $x$ and $y$. On the other hand, if $2M<2^{n+m}$, the denominator is negative, and so $|\alpha|^2\geq 0$ implies $M\leq 0$. This can only happen if $M=0$, i.e. all answers $F(x,y)$ give answer 1.
We conclude that unless $F(x,y)$ is constant, there is no valid $\alpha,\beta$ so that the measurement gives probability 1, which means there is no pure state description of that qubit.
