Nice question. I think that the answer is yes and more general. I have never seen this but seems natural.

${\bf Lemma}$ Let $f\colon \mathbb{P}^n_{\mathbb{C}}\to \mathbb{P}^n_{\mathbb{C}}$ be a rational map of degree $d$, given by $(x_0:\dots:x_n)\to (f_0:\dots:f_n)$, where the $f_i$ are homogeneous of degree $d$.

Suppose that the hypersurface $H\subset \mathbb{P}^n_{\mathbb{C}}$ given $f_0=0$ is irreducible and that the map from $H$ to $\mathbb{P}^{n-1}_{\mathbb{C}}$ given by the restriction (i.e. $(x_0:\dots:x_n)\to (f_1:\dots:f_n)$ on $H$) is birational.

Then, the map $f$ is birational.

${\bf Proof}$ Let $\sigma$ be the linear system associated, which corresponds to hypersurfaces of $\mathbb{P}^n$ of equation $\sum_{i=0}^n a_i f_i=0$. The restriction being birational, the intersection of $n-1$ general elements of $\sigma$ with $H$ give one mobile point, and a non-mobile part that we call $R$. Since every member of $\sigma$ contains $R$ and because elements have all the same degree, the intersection of $n$ general elements of $\sigma$ gives $R$, plus exactly one mobile point (the last part follows from Bézout). This shows that $f$ is birational.

${\bf Remark:}$ We could also view $R$ as a set of points, curves,... with some multiplicities and use intersection form on the blow-up.