No, your process does not make any sense. Even if the mean of $X_k$ is $1$, this does not mean that $$\frac {X_1 + \dots + X_N}{N} = 1$$
for some $N$.
This is a huge misunderstanding; I'll try to briefly explain why is it so.
Random variables are not numbers
When talking about a random variable, you're talking about a function. When you write $X_1$, what you mean is $X_1(\omega)$; that is, it takes a different value for different $\omega$. As such, you cannot expect $X_1(\omega) + \dots + X_N(\omega) = N(\omega)$ to hold! The left hand side depends on the particular $\omega$, as does the right hand side. Since they are independent, this equation cannot hold for any $N$ (random variable or not)
Relationship between mean and expectation
Your confusion probably stems from the fact that if $x_i$ are numbers, then we call their "mean" (or more properly average) the number $\frac 1n\sum_{i=1}^n x_i $. Clearly you can do the same with random variables (and the result will again be a random variable, not a number) to get to
$$\frac 1n\sum_{i=1}^n X_i(\omega)$$
(note that in this context I am assuming $n$ constant, different than $N(\omega)$ random variable)
A priori, this value has no relationship whatsover with the abstract integral $E[X_1]$ (which is what we call the mean of the random variable). They are completely different things defined in completely different ways! For instance, $E[X_1]$ only depends on $X_1$ while the other is an average of many random variables.
One discovers, after some work, that if the random variables $X_i$ are independent and indentically distributed with mean $\mu \in \mathbb R$, then $$\frac 1n \sum_{i=1}^n X_i \to \mu$$
(the limit is taken a.s. and in $L^1$).
So you if you take a lot of iid random variables and you take their average, the result is still a random variable, but it's "very close" to a constant, and that constant is precisely the mean $\mu$. As $n \to \infty$, it will be "equal" to the constant $\mu$. This is called the Law of Large Numbers and it's the justification as of why we call $E[X_1]$ "mean" in the first place, but you should remember how things are defined and not get confused by words
How to solve the problem
Now, if instead of $N(\omega)$ we simply had $n$, the problem was easy; use linearity of expectation to find that
$$E(X_1 + \dots + X_n) = E(X_1) + \dots + E(X_n) = n$$
If $N(\omega)$ is a random variable, clearly you cannot do the same; you need to use the tower property of conditional expectation to write that
$$E(X_1 + \dots + X_N) = E(E(X_1 + \dots + X_n \mid N = n) ) = E(N) = \sum_{k=1}^\infty \frac k{2^k} = 2$$