Even in finite dimensions you can easily change the inner product. Let $\{ e_{n} \}$ be an orthonormal basis of $\mathbb{C}^{N}$ and define the new inner product
$$
(x,y)_{\mbox{new}}=\sum_{n=1}^{N}\lambda_{n}(x,e_{n})(e_{n},y).
$$
where $\lambda_{n} > 0$ for all $n$.
All the norms are equivalent on $\mathbb{C}^{N}$. This can be written as
$$
(x,y)_{\mbox{new}}=(Ax,y)_{\mbox{old}}
$$
where $A$ is a positive definite selfadjoint matrix. In finite dimensions, this describes every possible inner-product. The inner products are in one-to-one correspondence with positive definite matrices. Because selfadjoint $A$ can be diagonalized by an orthonormal basis of eigenvectors, $(Ax,y) = \sum_{n}\lambda_{n}x_{n}y_{n}^{\star}$ always looks like a weighted inner product when viewed with respect to a correctly chosen orthonormal basis.
If $X$ is an infinite dimensional linear space on which two topologically equivalent Hilbert inner products are defined, say $(\cdot,\cdot)$ and $(\cdot,\cdot)_{1}$, then the same thing happens. There exists a unique positive bounded selfadjoint $A$ such that
$$
(x,y)_{1}=(Ax,y),\;\;\; x,y\in X.
$$
But it also goes the other way: $(x,y)=(Bx,y)_{1}$ where $B$ is positive. You end up with $(x,y)=(Bx,y)_{1}=(ABx,y)$ which gives $AB=I$. Similarly $BA=I$. The existence of such $A$ and $B$ comes from the Lax-Milgram Theorem, which is proved using the Riesz Representation Theorem for bounded linear functionals on a Hilbert Space.
But there are bounded positive linear operators $A$ on a Hilbert space $X$ which are not positive definite. Such an $A$ gives rise to $(x,y)_{1}=(Ax,y)$ with $\|x\|_{1} \le C\|x\|$ for a constant $C$, but the reverse inequality need not hold, which can lead to an incomplete $X$ under $\|\cdot\|_{1}$. For example, let $X=L^{2}[0,1]$ with the usual inner product. Define a new inner product by $(f,g)_{1}=\int_{0}^{1}xf(x)g(x)\,dx$. This is achieved as $(Af,g)$ where $Af=xf(x)$. This space is not complete because $1/\sqrt{x}$ is in the completion of $L^{2}$ under the norm $\|\cdot\|_{1}$. The completion of $(X,\|\cdot\|_{1})$ consists of $\frac{1}{\sqrt{x}}L^{2}[0,1]$. However, if you instead define $\|f\|_{1}^{2}=\int_{0}^{1}(x+\epsilon)|f(x)|^{2}\,dx$ for some $\epsilon > 0$, then you end up with an equivalent norm on $X$.
Since the author seems to make the statement so early in his book, it's probably safe to say that your intuition is good enough.
However, here is a way to make it precise: I would interpret "$S$ is smooth" as "$S$ is a smooth submanifold". If $S$ is the unit sphere with respect to a norm, then this would be equivalent to the following statement:
($\star$) There is a smooth function $F:\mathbb{R}^n \rightarrow \mathbb{R}$ with $S=\{x\vert F(x) = 1\}$ and $\nabla F(x)\neq 0$ for all $x\in S$.
Now if $S$ is the unit sphere with respect to an inner product $\langle \cdot , \cdot \rangle$, then you can take $F(x) := \langle x , x \rangle$.
It's easy to see that this is smooth (at least away from $0$, which is enough) and satisfies $\nabla F(x) = 2x$, hence ($\star$) is true.
If $S$ is the unit sphere with respect to the maximum norm, then ($\star$) is not satisfied. First note that $F(x) = \Vert x \Vert_{\max}^2$ is not a
smooth function. It might however be possible that another choice for $F$ works. But this is not possible: Note that $S$ is now a cube: Let $x_0$ be one of the corners and take two sequence $a_n,b_n\in S$ which approach $x_0$ from different faces. Since $\nabla F(x) \perp S$ (with respect to the usual dot-product) for all $x\in S$, we get $$\nabla F(x_0)= \lim \nabla F(a_n) \perp \lim \nabla F(b_n) = \nabla F(x_0)$$
and hence $\nabla F(x_0)= 0$, a contradiction.
Best Answer
What you can show is that the unit ball of any finite dimensional inner product space is an ellipsoid. That is, there exists a linear, invertible transformation $R$ such that the image of the unit ball is a sphere.
To see this, note that any inner product on $\mathbb R^n$ is of the form $\langle x,y\rangle = x^T A y$ for some positive definite matrix $A$. So apply the Cholesky factorization $A = R^T R$ to see that $\langle x,y\rangle = (Rx)^T (Ry)$. Then the image under $R$ of the unit ball induced by $\langle \cdot,\cdot\rangle$ is a sphere.
Another way to understand that not all norms come from an inner product is that any norm that comes from an inner product satisfies the polarization identity, that is $$ \|x\|^2 + \|y\|^2 = \tfrac12(\|x+y\|^2 + \|x-y\|^2) ,$$ and it can be shown that this is an if and only if condition. So just show that the max norm, or the taxi-cab norm fails this condition.