Wikipedia says "Let R be the set of all sets that are not members of themselves. If R is not a member of itself, then its definition dictates that it must contain itself, and if it contains itself, then it contradicts its own definition as the set of all sets that are not members of themselves."
However, let's use English linguistic history to provide a solution to Russell's paradox. Imagine an apple. I can say "do you want a small part or the whole part of the apple?" Anything is part of itself because you can have a whole part. Now let's say we have a treasure chest. Inside the treasure chest is treasure. If I asked someone "what does the treasure chest contain," they would say treasure. If I asked them "does the treasure chest contain a treasure chest," they would say no.
So anything is part of itself even though it does not contain itself. Anything that is not part of itself can contain itself (or else what else would contain itself?).
Let's look at Russell's paradox again. If R is not a member of itself, then that means it must not be contained by itself (it's not part of itself). It does not mean (at least not yet) that R contains itself because that would require a leap in logic. R is the set of all sets that are not members of itself, so the first step you would do is recognize that R must not be contained by itself, then you recognize that the R being contained and the R from the definition is the same. Now let's look at our statement about the treasure chest above. If anything does not contain itself, then it is part of itself. We know that R is not contained by itself and R = R, therefore R can contain itself. But R being the set of all sets that are not members of themselves does not mean that it's an element of itself. So R contains itself, but it's not an element of itself. There is no contradiction.
Wikipedia says R E R <==> R not(E) R, but R E R is incorrect. It should be R (backwards E) R.
Basically what I'm trying to say is containing itself and being contained isn't commutative as illustrated in the treasure chest example.
If we look at the barber paradox (a real world example of Russell's paradox), we have to look at it as a function of time. If "all those who do not shave themselves" includes the barber, then the barber is not necessarily one of them and he can shave himself. It's not commutative. At the instant he makes the decision to shave himself, he is no longer one of "all those who do not shave themselves," but the group still includes him. Whenever the barber is doing an action, at that exact instant, he is not part of any group involving actions except himself. It's impossible to do an action at the exact same time as someone else if time were divisible into infinite intervals. The barber is also shaving members of the group (not the whole group at the same time), so he will shave himself since he is not one of them (even though the group includes him).
In the barber example, once time passes, the situation will sort itself out so the barber will eventually be out of the group of "all those who do not shave themselves." We are just talking about the instant the barber makes his first decision of whether he should shave himself.
So is this a possible solution for Russell's paradox?
With these kinds of linguistic contortions you may be able to resolve Russell's paradox for sentences written in ordinary English. (I haven't read your argument in detail.)
But that does not help with Russell's paradox in formal mathematics. If your axiom system allows you to form the set $$ \{ x \ | \ x \not\in x\} $$ then you are in trouble.
Mathematicians have worked hard to write axiom systems that don't permit that but do permit what they need to do useful mathematics.