Suppose $\left(\phi,\phi^{\#}\right):X\longrightarrow Y$ is a morphism of schemes with the following properties:
1.$\phi$ is an isomorphism of topological spaces;
2.Every $\phi^{\#}_U:\mathcal{O}_Y(U)\longrightarrow \mathcal{O}_X(\phi^{-1}(U))$ is invertible
I'd like to say that then $(\phi,\phi^{\#})$ is an isomorphism of schemes. Unfortunately, I seem to always get confused by the technicalities of scheme theory, so I'm asking if my proof is good enough.
We must show an inverse of $(\phi,\phi^{\#})$. The inverse $(\varphi,\varphi^{\#})$ is defined as following
1.$\varphi=\phi^{-1}:Y\longrightarrow X$;
2.$\varphi^{\#}_V:\mathcal{O}_X(V)\longrightarrow \mathcal{O}_Y(\varphi^{-1}(V))$ is the inverse of $\phi^{\#}:\mathcal{O}_X(\varphi^{-1}(V))\longrightarrow \mathcal{O}_Y(V)$.
the first definition and the fact that $\phi$ is an isomorphism imply that $\varphi^{-1}(V)$ is open and that $\phi^{-1}(\varphi^{-1}(V))=V$.
No we need only to show that the $\varphi^{\#}_V$ are morphisms of sheaves, that is, they commute with the restrictions. Suppose $W\subseteq V$ and $s\in \mathcal{O}_X(V)$. Then $\varphi^{\#}_W(s|_W)=t$ where $t$ is the only element in $\varphi^{-1}(W)$ for which $\phi^{\#}(t)=s|_W$. On the other side, $\varphi^{\#}(s)=z$ has the property $\phi^{\#}(z)=s$ and $\phi^{\#}(z|_{\varphi^{-1}(W)})=s|_{W}$, so $$\varphi^{\#}(s)|_{\varphi^{-1}(W)}=t=\varphi^{\#}_W(s|_W)$$
Presumably your definition of an isomorphism of schemes is a morphism that has an inverse. But the notion of an inverse is defined in terms of composition of morphisms of schemes. If you have not already done so, it is a worthwhile exercise to carefully write down what the definition of composition of such morphisms is (this is not as trivial as it may appear at first glance). Whether you look it up somewhere or work it out for yourself, knowing how composition is defined is necessary for checking that any candidate for an inverse of your morphism really is an inverse (and this is something you have overlooked in your reasoning). If you would like me to include the details of this, I can, but for now I am going to refrain from doing so.
Anyway, you have assumed that $\phi$ is a homeomorphism and that for each open subset $U$ of $Y$, the ring homomorphism $\phi_V^\sharp:\mathscr{O}_Y(U)\to\mathscr{O}_X(\phi^{-1}(U))$ is an isomorphism (this second assumption is equivalent to the assertion that $\phi^\sharp:\mathscr{O}_Y\to\phi_*\mathscr{O}_X$ is an isomorphism of sheaves of rings). You want to define a morphism $(\psi,\psi^\sharp):Y\to X$ that serves as an inverse of $(\phi,\phi^\sharp):X\to Y$ (using $\varphi$ for the inverse is too confusing for me, so I am using $\psi$). Your definitions of $\psi$ and $\psi^\sharp$ are the natural (correct) ones. I think your verification that $\psi^\sharp$ is a morphism of sheaves is all right, but I personally find it hard to follow, and feel that you could clarify things with a bit more notation. Your definition in (2) amounts to setting, for each open subset $V$ of $X$, $\psi_V^\sharp=(\phi_{\phi(V)}^\sharp)^{-1}$. Now the commutativity formula you need to verify (expressing what is intended by compatibility with restriction) is
$$\mathrm{res}_{\phi(V),\phi(W)}\circ(\phi_{\phi(V)}^\sharp)^{-1} =(\phi_{\phi(W)}^\sharp)^{-1}\circ\mathrm{res}_{V,W}.$$
But this formula is equivalent to the formula
$$\phi_{\phi(W)}^\sharp\circ \mathrm{res}_{\phi(V),\phi(W)} =\mathrm{res}_{V,W}\circ\phi_{\phi(V)}^\sharp.$$
The second formula is a particular instance of the compatibility of $\phi^\sharp$ with restriction, so it holds, which means the first formula holds as well. In this way you make a bit clearer how $\psi^\sharp$ inherits this compatibility from that of $\phi^\sharp$.
As I said earlier, you still need to confirm that $(\psi,\psi^\sharp)$ really is an inverse of $(\phi,\phi^\sharp)$, which is not difficult, but does require you to know precisely how to compose two morphisms of schemes. If you know the formula for the sheaf map of a composition, then you can use the alternative but equivalent characterization of $\psi^\sharp$ as $\psi_*(\phi^\sharp)^{-1}$ to make short work of the necessary computations (but you can also do them open-set-by-open-set).