Let $\varphi\colon \mathbb{R}^2\rightarrow \mathbb{R}^3, \ \varphi(u,v)=\begin{pmatrix} u\\ v^3(1+v^3)+\sinh(u)\\ v^3(1-v^3) -\sinh(u)\end{pmatrix}$
I want to find and inverse of $\varphi$ on its image $\varphi(\mathbb{R}^2)$.
So be $\varphi(u,v)=(x,y,z)^t$. We then immediately see that for every $x\in\varphi(\mathbb{R}^2)$ we get a $u=x$.
We only miss a $v$ and therefore compute $y+z=2v^3$ and arrive at $\sqrt[3]{\frac{y+z}{2}}=v$.
The only problem is, $y+z$ must not be positive, so $v$ might not be defined.
Obviously not every $(x,y,z)^T$ can be inverted. For example $(0,1,1)^T$. From your equations you would get $u=0$ and $v=1$, but $\varphi(0,1)=(0,2,0)^T$. That's because you get three equations with only two unknowns. So the way I would approach this problem is I choose only two of the components in the image of $\varphi$ in $\mathbb R^3$, and write the third one in terms of those two. Given $y$ and $z$, you calculated $2v^3=y+z$. Now plug this into the equation of $y$: $$\frac{y+z}{2}\left(1+\frac{y+z}{2}\right)+\sinh x=y$$ or $$\sinh x=y-\frac{y+z}{2}\left(1+\frac{y+z}{2}\right)$$ Note that if you plug in this value in the equation for $z$ you get an identity. Then the image of $\varphi$ is the set$$\DeclareMathOperator\arcsinh{arcsinh}\begin{pmatrix}\arcsinh\left(y-\frac{y+z}{2}\left(1+\frac{y+z}{2}\right)\right)\\y\\z \end{pmatrix}$$ This yields $$\varphi^{-1}\begin{pmatrix}\arcsinh\left(y-\frac{y+z}{2}\left(1+\frac{y+z}{2}\right)\right)\\y\\z \end{pmatrix}=\begin{pmatrix}u\\v\end{pmatrix}$$ with $v=\sqrt[3]{\frac{y+z}{2}}$ and $u=\arcsinh\left(y-\frac{y+z}{2}\left(1+\frac{y+z}{2}\right)\right)$. If you want, you can further simplify $u$.