I do not think that there exists any mixed strategy asymmetric equilibria with two players.
The trick is always the same when looking for mixed strategy equilibria :
if a player plays different pure strategies with strictly positive probability at an equilibrium, these pure strategies must yield the same expected payoff given how the other player plays.
Now, I am not 100% positive how this applies to cases with a continuum of possible strategies, but I am pretty confident one can prove that "playing different pure strategies with positive probability" simply extends to "playing different pure strategies with positive probability mass". Then you can see that the only way for player $1$ to put strictly positive probability mass on two different pure strategies while being at an equilibrium is for player $2$ to play a uniform distribution over $[0;1]$.
Regarding the existence of asymmetric equilibria with more bidders, I do not think there exists any either, at least when players play continuous strategies. This time, for any player $i$, the relevant information regarding other players action is only what is the maximum of others bids (as we assumed away discrete strategies). So for $i$ to play different pure strategies with strictly positive probability, it needs to be the case that
$$ \max \{ b_1, \dots , b_{i-1}, b_{i+1} ,\dots b_n\} \sim U[0,1] $$
which is equivalent to
$$ P(b_1 \leq p, \dots , b_{i-1} \leq p, b_{i+1}\leq p ,\dots b_n \leq p) = p $$
$$ F(b_1 \leq p) * \dots * F(b_{i-1} \leq p) * F(b_{i+1}\leq p) *\dots* F(b_n \leq p) = p$$
This work for every $i$ if $F(b_j \leq p) = p^{1/n-1}$ for all $j \in \{1,\dots,n\}$, but I do not see how it could work with asymmetric bidding function. In effect, assume that $F_j \neq F_k$. We must have
$$ F_j \prod_{i \neq j,k} F_i = p$$.
and
$$ F_k \prod_{i \neq j,k} F_i = p $$
but this contradicts $F_j \neq F_k$.
You might still want to consider the possibility of there being asymmetric equilibria with discrete mixed strategies. The former argument does not apply to these cases, but my guess would be that there are none.
In a Nash equilibrium, no player has incentive to change their action, holding fixed the actions of the others. Here, actions are bids.
Take the action profile proposed by Osborne and Rubinstein. Does player one have incentive to increase bid? No, he will still win but pay more. Does he have incentive to lower it? No, he will lose the auction, and give up the surplus (at least zero) he is current receiving.
Do other's have incentive to lower their bids? No, they will continue to lose the auction. Do others have incentive to increase their bids? No, either they will continue to lose the auction, or if they raise it enough, they will win, but at a price that is at least as high as their valuation since $b_1 \geq v_2$.
Now, we consider your proposed strategy: player 1 pays $v_2$ and every other players bid whatever they want $b_i ≤ v_2$. This may not be a Nash equilibrium. If other's all bid zero for instance, than player 1 has incentive to lower his bid to zero, he will continue to win and pay nothing! In fact, if the other's leave player 1 and room to lower his bid, he will. That's why, in equilibrium we need someone else to bid what player one bids in equilibrium.
But, why are these the only equilibrium? Well, suppose someone else wins. If it is an equilibrium, they must have bid no more than their valuation. Otherwise, they would be better off losing the auction and so they would have incentive to bid zero. But, if the winning bidder isn't player 1 and is bidding a value no higher than their own valuation, player 1 has incentive to raise his bid to just above the currently winning bid and win the auction at a price that gives him a positive surplus. Thus, in equilibrium, player 1 has to win.
Best Answer
The notion "symmetric equilibrium" (the one from Wikipedia article) is not applicable here, because the game is not symmetric (different players have different "profits per click").
I've a look at the paper and I think, that "symmetric Nash equilibrium" in your case is nothing but technically convenient case of Nash equilibrium (the latter is not unique in your case).
Also I've noted a probable misprint in the proof of the "Fact 1". It should be $(v_s-p_s)x_s \geq (v_s-p_{S+1})x_{S+1}$ instead of $(v_s-p_s)x_s \geq (v_{S+1}-p_{S+1})x_{S+1}$.