First, a caveat: I'm on the job market this year in the midst of the couple weeks when calls are rolling in. Hence, this seemed like a nice way to kill some time (semi-)productively. This is also a disclaimer in case I've made an error :)
Now, let's look at the one you're suggested to try for, where player $1$ chooses $0$ with $p$ and $1$ with $1-p$. Again let player $2$ choose $G$ with full support.
If player $1$ chooses $0$ and player $2$ chooses $y$, Pl $1$ gets $$\int_{0}^{1}yg(y)dy = \mu$$
and if he chooses $1$, he gets $$\int_{0}^{1}(1-2y)g(y)dy = 1 - 2\mu$$
where $\mu := \mathbb{E}_{G}[y]$
Hence, for indifference we need $\mu = 1/3$. Ah but we also need this to be robust to any other deviation. Namely, we need.
$$\frac{1}{3} \geq x\int_{0}^{x}(1-2y)g(y) + (1-2x)\int_{x}^{1}yg(y)$$
for all $x \in [0,1]$.
Knowing that $\mathbb{E}_{G}(y) = 1/3$ for $G(y) = \sqrt{y}$, let's guess that that is the solution. The right-hand side reduces to
$$\dfrac{2x^\frac{3}{2}-2x+1}{3}$$ which is maximized at $x = 0$ and $x = 1$ and equals $1/3$ there, as required.
Finally, for player $2$, she is indifferent over any $y$ if $$p(-y) - (1-p)(1-2y)$$ does not depend on $y$. This holds if $p = 2/3$.
Thus the equilibrium is given as follows: player $1$ chooses $0$ with probability $2/3$ and $1$ with probability $1/3$; and player $2$ chooses cdf $G(y) = \sqrt{y}$ on $[0,1]$. The value of the game is $1/3$ for player $1$ and $-1/3$ for player $2$.
