The 1-dimensional Heisenberg group $\mathbb{H}^1$ with group operation $$(s,x,y)\cdot(s',x',y') = (s+s' + \frac{1}{2}(xy' - x'y), x+x', y+y')$$ is supposedly isomorphic to $\mathbb{R}^3$ with group law given by $$(s,x,y)\star(s',x',y') = (s+s' + xy', x+x', y+y')$$ by using the map (call it $f$) $(s,x,y) \mapsto (s+\frac{1}{2}xy, x,y)$. How exactly do we show this with computation? I get extra terms that shouldn't be there when I try myself.
So checking the relation $f((s,x,y)\cdot (s',x',y')) = f((s,x,y))\star f((s',x',y'))$, beginning with the LHS:
$$f((s,x,y)\cdot (s',x',y')) = f((s+s' +\frac{1}{2}(xy' - x'y),x+x',y+y')) = (s+s' +\frac{1}{2}(xy' - x'y) + \frac{1}{2}(x+x')(y+y'),x+x',y+y') = (s+s' + \frac{1}{2}(2xy' + xy + x'y'), x+x',y+y') $$.
Looking at the RHS now:
$$f((s,x,y))\star f((s',x',y')) = (s+\frac{1}{2}xy,x,y)\star (s'+\frac{1}{2}x'y', x',y') = (s+s'+\frac{1}{2}xy+ \frac{1}{2}x'y' + \frac{1}{2}xy',x)$$
As you can see I am off by a factor of a $\frac{1}{2} xy'$.
Suppose that $f:(s,x,y)\mapsto(s+c_1xy/2,x,y)$ is a mapping between the two group laws: $$(s,x,y)\cdot(s',x',y') = (s+s' + \frac12(c_2xy' +c_3 x'y), x+x', y+y') $$ and $$(s,x,y)\star(s',x',y') = (s+s'+\frac12((c_1+c_2)xy'+(c_1+c_3)x'y),x+x',y+y')$$ which are both associative. Then verify the equation $$f((s,x,y)\cdot(s',x',y'))=f(s,x,y)\star f(s',x',y').$$ In your case, you want to specialize to $\;c_1=c_2=1,c_3=-1.$