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.