rock, paper, scissors, well

You are quite right. Rock is dominated by Well: no matter what the opponent plays, you do at least as well playing Well as you would playing Rock, and against Rock or Well you do better. Good players will never play Rock, so the game reduces to Well, Paper, Scissors, which is isomorphic to Rock, Paper, Scissors.

Mixing evenly between paper, scissors, and well is indeed an equilibrium.

Starting with Vadim's condition:

$$ p-s+(1-r-p-s)=\\-r+s-(1-r-p-s)=\\r-p+(1-r-p-s)=\\-r+p-s$$

If Rock receives no weight, we have:

$$s-(1-p-s)=\\-p+(1-p-s)=\\p-s$$

Which gives $p=s=(1-p-s)=\frac{1}{3}$

Further, Rock is dominated by any combination of the other three strategies against this mixture. Thus, any mixture of the three is indeed a best response to an equal mixture and so the equal mixture is a Nash equilibrium.

To see there is no other equilibrium, we can use the fact that in a symmetric non-zero sum game, any strategy optimal for one player is optimal for another. Note that when rock receives weight in opponent's strategy, rock is strictly dominated by well. Thus, rock cannot be part of an equilibrium since it would imply that rock is part of an optimal strategy against a strategy that includes positive weight on rock.

Here is the payoff matrix: $$\begin{matrix} &R&P&S&W\\ R&0&1&-1&1\\ P& -1&0&1&-1\\ S & 1 & -1 & 0 & 1\\W & -1 & 1 & -1 & 0\end{matrix}$$

Suppose we pick $R$ with probability $r$, $P$ with probability $p$, $S$ with probability $s$, and $W$ with probability $1-r-p-s$. An optimal mixed strategy is indifferent to the opponent's choice. Hence $$ p-s+(1-r-p-s)=\\-r+s-(1-r-p-s)=\\r-p+(1-r-p-s)=\\-r+p-s$$

Unfortunately there is no solution to this. Alas, I don't have time right now to take this further.