We can change $s^2 = \frac{t^2 - 1}{2t}$ to $$2ts^2 = t^2 - 1$$ $$16t^2s^2 = 8t^3 - 8t$$ and making the change of variables $(x, y) = (2t, 4ts)$, we get the elliptic curve $y^2 = x^3 - 4x$
Can we do something similar with $s^2 = \frac{t^2 + 2t -1 }{t^2 - 2t - 1}$?
Any insight is appreciated.
We can do something similar to what you did with your example. Cross-multiplying, and then making various adjustments, gives that
$$\begin{equation}\begin{aligned} s^2(t^2 - 2t - 1) & = t^2 + 2t - 1 \\ s^2(t^2 - 2t - 1)^{2}t^2 & = ((t^2 - 1) + 2t)((t^2 - 1) - 2t)t^2 \\ (s(t^3 - 2t^2 - t))^2 & = ((t^2 - 1)^2 - 4t^2)t^2 \\ & = ((t^4 - 2t^2 + 1) - 4t^2)t^2 \\ & = t^6 - 6t^4 + t^2 \\ & = (t^6 - 6t^4 + 12t^2 - 8) - 12t^2 + 8 + t^2 \\ & = (t^2 - 2)^3 - 11t^2 + 8 \\ & = (t^2 - 2)^3 - 11(t^2 - 2) - 22 + 8 \\ & = (t^2 - 2)^3 - 11(t^2 - 2) - 14 \end{aligned}\end{equation}$$
Thus, using the change of variables $(x, y) = (t^2 - 2, s(t^3 - 2t^2 - t))$, we then have a similar elliptic curve to your example, in particular we get
$$y^2 = x^3 - 11x - 14$$
Note that, as explained in Wikipedia's Elliptic curve article, the general form of $y^2 = x^3 + ax + b$ requires the curve to be non-singular (i.e., square-free in $x$), which applies here since
$$4a^3 + 27b^2 = 4(-11)^3 + 27(-14)^2 = -32 \neq 0$$
In particular, the cubic's roots are $-2$ and $1 \pm 2\sqrt{2}$.
Also, as indicated in Jyrki Lahtonen's comment, note we can generally reverse the mapping, i.e., write $(s, t)$ in terms of $(x, y)$. In particular, there's
$$x = t^2 - 2 \;\;\to\;\; t = \pm\sqrt{x + 2}$$
If $t^3 - 2t^2 - t = 0$, then $y = 0$ and $s$ is indeterminate, else we get
$$s = \frac{y}{t^3 - 2t^2 - t} = \frac{y}{\pm(\sqrt{x + 2})^3 - 2(x + 2) \mp\sqrt{x + 2}}$$
However, as stated in Jyrki Lahtonen's other comment, involving square roots means the mapping is not rational. Thus, what I've described above is an isogeny to the elliptic curve $y^2 = x^2 - 11x - 14$ rather than being an isomorphism.
It is not quite clear to me what you need/want. I am guessing that you seek to find an elliptic curve that shares the same function field. But, you did not disclose the field of constants either. Anyway, I followed the procedure in this answer by David E. Speyer as an exercise for myself, so I might as well share.
The transformation is done in several steps. The end result is an elliptic curve defined over $\Bbb{Q}$ with the property that the function field $\Bbb{Q}(s,t), s^2=(t^2+2t-1)/(t^2-2t-1)$ is equal to $\Bbb{Q}(x,y), y^2=x^3-x$.
Like John Omielan I begin by replacing $s$ with $u=s(t^2-2t-1)$ and arrive at the polynomial equation $$ u^2=(t^2-2t-1)(t^2+2t-1)=t^4-6t^2+1.\tag{1} $$ It is known that the polynomial equation $y^2=f(x)$, $f$ squarefree, defines a curve of genus $g=(m-2)/2$ when $m=\deg f$ is even, but $g=(m-1)/2$ when $m$ is odd. So here we can already see that your equation defines a genus one curve. It obviously has rational points, so it will be an elliptic curve.
The equation in the linked thread is very similar (biquadratic!) so all I needed to do was to modify the constants a bit. So I shamelessly replace $u$ and $t$ with $v$ and $w$ defined by the equations $$u=t^2-3+v,\qquad t=w/v. $$ Observe that $v=(u-t^2+3)$, $w=t(u-t^2+3)$, so this change of variables is birational. Plugging these into $(1)$ and clearing the numerators yields the equation $$ 8 v - 6 v^2 + v^3 + 2 w^2=0,\tag{2} $$ from which we already see that it, indeed, defines an elliptic curve.
It may be worth our while to find the minimal Weierstrass model. There isn't much to this. Looking at $(2)$ we see that $w=2y$, $v=-2x+A$ is strongly suggested. The choice $A=2$ gets rid of the quadratic term, so I plug in $w=2y$, $v=-2x+2$ into $(2)$, and divide the end result by eight: $$ y^2=x^3-x.\tag{3} $$
Consider the curve $$y^2 = \frac{(x-\alpha)(x-\beta)}{(x-\gamma)(x-\delta)}$$
Write $$y =\frac{x-\delta}{x-\gamma}\cdot y_1$$ so
$$y_1^2 =\frac{(x-\alpha)(x-\beta)(x-\gamma)}{(x-\delta)^3}= \frac{P_3((x-\delta))}{(x-\delta)^3}=Q_3(\frac{1}{x-\delta})$$
($P_3$ is a cubic polynomial, $Q_3$ its reverse).
