Remember that a basis of a vector space only needs to (1) span the vector space, and (2) be linearly independent. In particular, a basis does not have to be orthogonal, and it certainly doesn't have to be normalized. And one of the most common types of basis (a coordinate basis) is usually not normalized.
You're confused because you usually see the metric tensor in spherical coordinates given as
\begin{equation}
\mathbf{g} =
\begin{pmatrix}
1 & 0 & 0 \\
0 & r^2 & 0 \\
0 & 0 & r^2 \sin^2\theta
\end{pmatrix}.
\end{equation}
This is the metric with respect to the coordinate basis, whereas you've (correctly) written the metric with respect to the orthonormalized vector basis — and it's very important to remember the distinction between those types of bases. I'll explain.
Let's write the coordinate basis vectors as
\begin{equation}
\mathbf{r}, \boldsymbol{\theta}, \boldsymbol{\phi}.
\end{equation}
(Note that I'm using a bold font to indicate that these are vectors, but I'm not putting hats on them, for reasons that will become clear soon.) These vectors represent the amount you would move through the space if you changed the corresponding coordinate by a certain amount. For example, if $\mathbf{p}(r, \theta, \phi)$ is the position vector to the point with spherical coordinates $r, \theta, \phi$, then those coordinate basis vectors are defined as
\begin{align}
\mathbf{r} &= \frac{\partial \mathbf{p}} {\partial r} \\
\boldsymbol{\theta} &= \frac{\partial \mathbf{p}} {\partial \theta} \\
\boldsymbol{\phi} &= \frac{\partial \mathbf{p}} {\partial \phi}.
\end{align}
To relate that back to your basis given in Cartesian components, remember that
\begin{equation}
\mathbf{p} = r\sin\theta\cos\phi\, \hat{\mathbf{x}} + r\sin\theta\sin\phi\, \hat{\mathbf{y}} + r\cos\theta\, \hat{\mathbf{z}},
\end{equation}
which we can differentiate to find
\begin{align}
\mathbf{r} &= \sin\theta\cos\phi\, \hat{\mathbf{x}} + \sin\theta\sin\phi\, \hat{\mathbf{y}} + \cos\theta\, \hat{\mathbf{z}} \\
\boldsymbol{\theta} &= r\cos\theta\cos\phi\, \hat{\mathbf{x}} + r\cos\theta\sin\phi\, \hat{\mathbf{y}} - r\sin\theta\, \hat{\mathbf{z}} \\
\boldsymbol{\phi} &= -r\sin\theta\sin\phi\, \hat{\mathbf{x}} + r\sin\theta\cos\phi\, \hat{\mathbf{y}}.
\end{align}
Using these expressions, it's a simple exercise to see that we have
\begin{align}
\mathbf{r} \cdot \mathbf{r} &= 1 \\
\boldsymbol{\theta} \cdot \boldsymbol{\theta} &= r^2 \\
\boldsymbol{\phi} \cdot \boldsymbol{\phi} &= r^2 \sin^2 \theta.
\end{align}
So this basis is not orthonormal — and that's where the "usual" metric components come from, which is why the metric isn't just the identity as you expected. In fact, usually the only type of coordinates that lead to orthonormal basis vectors is a Cartesian coordinate systems (though even Cartesian coordinates are not orthonormal in nontrivial geometries).
On the other hand, a nearly identical simple exercise shows that your basis $(\mathbf{e}_r, \mathbf{e}_\theta, \mathbf{e}_\phi)$ is orthonormal. In fact, comparing our expressions in the Cartesian basis, we see that
\begin{align}
\mathbf{r} &= \mathbf{e}_r \\
\boldsymbol{\theta} &= r\, \mathbf{e}_\theta \\
\boldsymbol{\phi} &= r\sin\theta\, \mathbf{e}_\phi.
\end{align}
In an orthonormal basis, the metric is — essentially by definition — just the identity matrix, which is what you found.
Best Answer
This properly belongs on math.se, but to properly derive these you need to remember that we can write a vector in terms of basis vectors. The vector $\vec{A}$ is unchanged, but it is just expressed as a different linear combination: $$\vec{A} = A_x \hat {x} + A_y \hat{y} = A_r \hat{r} + A_\theta\hat{\theta} $$.
Because you can write $\hat{r}$ as a linear combination of $\hat x$ and $\hat y$, i.e., $\hat {r} = \frac{x}{r} \hat {x} + \frac{y}{r} \hat y = \cos\theta \hat{x} + \sin\theta \hat{y}$, and similarly for $\hat{\theta} = -\sin\theta \hat{x} + \cos\theta \hat y$, you can solve for the $A_r$ and $A_\theta$ in terms of $A_x$ and $A_y$.