In undergrad, I remember computing the group law of the nodal cubic $y^2= x^3 + x^2$ using a particularly slick parametrization. The usual parametrization of the nodal cubic is $(t^2-1, t^3-t)$, and if you use this parametrization, and solve for the relationship between three collinear points on the cubic, you end up getting that if $(t^2-1, t^3 -t)$, $(s^2-1, s^3 -s)$, and $(u^2-1, u^3 -u)$, are collinear, then $$u = - \frac{1+st}{(s+t)}.$$
This is done by simply writing down the slope of the line going through the three points in two different ways, and then factoring out the degenerate cases $s = t$, $s=u$ and $t=u$.
Unfortunately, from this parametrization and solution makes it hard to see what the group law on the nodal cubic actually is. There is the pesky denominator and plus one term lurking around. I'm totally blanking on the right strategy for reparametrizing to make the equation nicer.
(Bonus points if you can help me do the same with the cuspidal cubic $y^2 = x^3$. The parametrization $(t^2, t^3)$ gives the formula for a trio of collinear points $(t^2, t^3), (s^2, s^3),$ and $$\left(\frac{(st)^2}{(s+t)^2}, -\frac{(st)^3}{(s+t)^3}\right) $$ which is, again, less than ideal.)
(I assume the characteristic is not $2$)
The problem with your parametrisation is that the set of smooth points of the curve ends up being $\Bbb P^1(K) \setminus \{-1;+1\}$. Do you know any natural group law on $\Bbb P^1(K) \setminus \{-1;+1\}$ ? I don't either.
A good thing to do is to move $\pm 1$ to a better pair of points, like $0$ and $\infty$. Do you know any natural group law on $\Bbb P^1(K) \setminus \{0,\infty\} = K^*$ ? Yes ? Perfect ! While we are at it we can also move $\infty$ to $1$ if we want to choose that point for the origin.
Let $f(x) = \frac {x+1}{x-1}$, and call $s',t',u'$ the inverse images of $s,t,u$ by $f$. (this $f$ was chosen so that $f(0)=-1, f(\infty)=+1, f(1)=\infty$)
Now, you said $s,t,u$ are aligned if and only if $st+su+tu+1=0$.
Replacing $s$ with $f(s')$ and so on you get the equivalent equation
$(s'+1)(t'+1)(u'-1)+(s'+1)(t'-1)(u'+1)+(s'-1)(t'+1)(u'+1)+(s'-1)(t'-1)(u'-1) = 0$
which simplifies to $4s't'u' = 4$
Choosing $(u = \infty ; u' = 1)$ for the origin, the group law is then the usual multiplication.
As for your other curve, you want to use $f(x) = x^{-1}$ to put the singular point at $\infty$.
The initial equation is $st+su+tu = 0$, and you end up with $u'+t'+s' = 0$, so you recover the regular addition on $K$.