It's easy to check that $a=0,\dfrac{\sqrt{2}}{2},-\dfrac{\sqrt{2}}{2}$ satisfy the condition.So we only need to prove no other solution exists. Suppose there is another solution $a$,assume $a>0$.If$a_n=0$ for some $n$,then $a_{n-1} \in \{-1,1\}$ a contradiction.So there is no 0 in the sequence.And we can get $$ a_{n+2} - a_n = \dfrac{1-2a_n^2}{a_n(a_n^2-1)} \cdots \cdots (1) $$for all $n$,where $a_n^2-1<0$. if$\dfrac{\sqrt{2}}{2}<a<1$,by (1) and induction we know that $\{a_n:n = 1,3,\cdots \}$ is increasing;if $0<a<\dfrac{\sqrt{2}}{2}$,notice that positive and negative number appear alternatively in $\{a_n\}$,similarly we can learn that $\{a_n:n = 1,3,\cdots \}$ is positive and have lower bound 0. But $\{a_n:n = 1,3,\cdots \}$ is bounded by -1 and 1,so $\lim_{n \to \infty} a_{2n-1} := A$ exists.Let $n \to \infty$ in (1) we get $\dfrac{1-2A^2}{A(A^2-1)}=0$,so $A = \dfrac{\sqrt{2}}{2}$,which satisfy neither of the two cases,a contradiction.

Here is a more direct solution. Assume that $a\ne 0$. Define $f(x) = x - \tfrac{1}{x}$ and define a sequence $b_n$ by $$b_1 = 1, b_n = \frac{\sqrt{b_{n-1}^2+4}-b_{n-1}}{2}\ (n=2,3,4,...)$$Observe that $b_{n-1} = -f(b_n)$. We claim that. Claim 1 : $b_{2n} < |a| < b_{2n+1}$ for every $n=0,1,2,...$. Proof : We use induction. Denote $P(n)$ by the assertion $$\begin{cases} |a| > b_{n} & n \text{ even} \\ |a| < b_{n} & n \text{ odd} \end{cases}$$Suppose that $P(n)$ is true. By induction hypothesis, we have $$ \begin{cases} |a_2| > b_{n} & n \text{ even} \\ |a_2| < b_{n} & n \text{ odd} \end{cases} \implies \begin{cases} |f(a)| > f(-b_{n+1}) & n \text{ even} \\ |f(a)| < -f(-b_{n+1}) & n \text{ odd} \end{cases} $$As $f$ is increasing, this yields the conclusion. We now investigate the limit of $b_n$. It turns out that both lower and upper bound converge to the same limit $\tfrac{1}{\sqrt{2}}$. Claim 2 : $b_n < \tfrac{1}{\sqrt{2}} \implies b_{n+1} > \tfrac{1}{\sqrt{2}}$ and $b_n > \tfrac{1}{\sqrt{2}} \implies b_{n+1} < \tfrac{1}{\sqrt{2}}.$ Proof : As $f$ is increasing and $f\left(\tfrac{1}{\sqrt{2}}\right) = -\tfrac{1}{\sqrt{2}}$, we get \begin{align*} b_n < \frac{1}{\sqrt{2}} &\implies f(-b_{n+1}) < f\left(-\frac{1}{\sqrt{2}}\right) &\implies b_{n+1} > \frac{1}{\sqrt{2}} \\ b_n > \frac{1}{\sqrt{2}} &\implies f(-b_{n+1}) > f\left(-\frac{1}{\sqrt{2}}\right) &\implies b_{n+1} < \frac{1}{\sqrt{2}} \\ \end{align*}implying the conclusion. Claim 3 : Sequence $\{b_{2n+1}\}$ is decreasing and sequence $\{b_{2n}\}$ is increasing. Proof : Notice that if $x = b_{n+2}$, then $$b_{n+2} - b_n = x + b_{n+1} - \frac{1}{b_{n+1}} = x + \frac{1-x^2}{x} - \frac{x}{1-x^2} = \frac{1-2x^2}{x(1-x^2)}.$$Hence $b_{n+2} > b_n \iff b_n < \tfrac{1}{\sqrt{2}}$ so combining with Claim 2 implies the conclusion. Claim 4 : $\lim_{n\to\infty} b_{2n+1} = \lim_{n\to\infty} b_{2n} = \tfrac{1}{\sqrt{2}}$. Proof : Since both sequences are monotone and bounded, by Monotone convergence theorem, both sequence converge. Let the limit be $s, t$ respectively. We see that $$s = \frac{1}{t} - s\text{ and } t = \frac{1}{s} - t$$Solving this system gives $s=t=\tfrac{1}{\sqrt{2}}$, implying the conclusion. By Claim 1 and Claim 4, $|a| = \frac{1}{\sqrt{2}}$. Hence the only possible solutions are $0, \pm\tfrac{1}{\sqrt{2}}$.