You can't embed a pseudo-Riemannian manifold into the Euclidean space or any Riemannian manifold at all. If you introduce some coordinates $y^a$ on the submanifold in the embedding manifold with metric $g_{\mu\nu}$ it gets the induced metric,
\begin{equation}
h_{ab}=\frac{\partial x^\mu}{\partial y^a}\frac{\partial x^\nu}{\partial y^b} g_{\mu\nu}
\end{equation}
where summation over the repeated indices is implied. Consider then the quadratic form on the two vectors on the submanifold,
\begin{equation}
(\xi,\zeta)=\xi^a h_{ab} \zeta^b
\end{equation}
From these two vectors on the submanifold we can construct two vectors in the embedding manifold such that,
\begin{equation}
\xi^\mu =\xi^a\frac{\partial x^\mu}{\partial y^a},\quad \zeta^\mu =\zeta^a\frac{\partial x^\mu}{\partial y^a}
\end{equation}
Then we may compute that quadratic form using the metric of the embedded space which is the main idea of the induced metric,
\begin{equation}
(\xi,\zeta)=\xi^\mu g_{\mu\nu} \zeta^\nu
\end{equation}
If the embedding space is Riemannian the rhs is positively definite and that means that $h_{ab}$ is also positively definite i.e. Riemannian.
If you consider the surface you defined embedded in the Euclidean space you don't get de Sitter or anti-de Sitter. You just get a hyperboloid with positively definite Riemannian metric. It also not a homogeneous space like (A)dS. The farther you go from its center the flatter the hyperboloid becomes. Whereas you can go to any point of (A)dS by performing a boost in the embedding space that leaves the surface as a whole invariant (just as a sphere in the Euclidean space is invariant under rotations). As result (A)dS is the spacetime of the constant curvature.
Of course you may say that your embedding of (A)dS does not relate its metric structure to the metric of the embedding space. But then your question is meaningless because the only difference between the Euclidean space and Minkowski spacetime is the metric. Without it both of them are just $\mathbb{R}^n$.
