I was trying to think of multiplicative vector spaces, with scalar multiplication replaced by exponentiation and I came up with these as examples
$C_p$ with $f: \mathbf{Z}_p\times C_p\to C_p$ given by $(n,r)\mapsto r^n$
$S^1$ with $f:\mathbb{R}\times S^1\to S^1$ given by $(a,z)\mapsto z^a$
$((F\setminus\{0\})^n, *,1)$ with normal exponentiation over $F$ where $F$ is a field.
I just wanted to check I was right in thinking these are indeed vector spaces?
In a vector space, if $\alpha$ is a scalar and $\mathbf{v}$ is a vector, then $\alpha\mathbf{v}=\mathbf{0}$ if and only if $\alpha = 0$ or $\mathbf{v}=\mathbf{0}$.
In your second example, the vector addition is supposed to be multiplication of elements of $S^1$, which means the zero vector is the number $1$; but if you take $\alpha=2\neq 0$ and $\mathbf{v}=-1\neq\mathbf{0}$, then you get $\alpha\cdot\mathbf{v} = (-1)^2 = 1 = \mathbf{0}$. Thus, the second example cannot be a vector space, and you messed up your verification somewhere. But since you don't tell us what you did, I don't know where you messed up.
The third example is not well-defined; not every field is closed under exponentiation, so you can't just say "standard exponentiation"; for instance, if $F=\mathbb{Q}$, then you cannot calculate $2^{1/2} = \frac{1}{2}\cdot 2$; in $\mathbb{R}$ you cannot take $(-1)^{1/2}$; and in $\mathbb{C}$ you run into a problem like the one mentioned above for the second example because exponentiation has a periodic component: if $\mathbf{v}=e$ and $\alpha=2\pi i$, then $\alpha\cdot\mathbf{v} = e^{\pi i} = \cos(2\pi) + i\sin(2\pi) = 1=\mathbf{0}$, even though $\mathbf{v}\neq\mathbf{0}$ and $\alpha\neq 0$. So the third example is not a vector space, and you messed up your verification somewhere. But since you don't tell us what you did, I don't know where exactly you messed up.
(It also goes wrong for finite fields, because the cardinality just doesn't work; a vector space of dimension $k$ over a field with $q$ elements has $q^k$ elements; but your underlying set here would have $(q-1)^k$ elements, which cannot be a power of $q$.)
The first example is fine, provided that you are viewing $C_p$ multiplicatively: if $C_p=\{1,x,x^2,\ldots,x^{p-1}\}$, and you look at $\mathbb{Z}_p$ as the field with $p$ elements, what you are really doing is looking at the standard $1$ dimensional vector space, except that, because instead of using addition for your vector operation you are using multiplication, then scalar multiplication becomes exponentiation: taking "logarithms base $x$" you obtain an isomorphism to $\mathbb{F}_p$ as a vector space over itself: if $\log_x(x^a) = a\bmod p$, we use $\oplus$ for the vector addition, and use $\odot$ for the scalar product you defined, we have: $$\begin{align*} \log_x(x^a\oplus x^b) &= \log_x(x^ax^b) = \log_x(x^{a+b}) = a+b = \log_x(x^a)+\log_x(x^b)\\ \log_x(r\odot (x^a)) &= \log_x((x^a)^r) = \log_x(x^{ar}) = ar = r\cdot \log_x(x^a). \end{align*}$$
What goes wrong in Example 2 is a bit subtle. Associativity doesn’t really work for the scalar multiplication.
Probably the best way to understand this is to really nail down the elements and vector addition; because otherwise, you run into issues with the scalar multiplication, e.g., is $\frac{1}{2}\cdot (-1)$ equal to $i$, or to $-i$? Is $\frac{1}{2}\cdot 1$ equal to $1$, or to $-1$? And once you nail it down, you’ll run into issues, precisely because this is not a continuous operation on $S^1$.
For instance: identify $S^1$ with the real numbers in $[0,2\pi)$ by writing $z=e^{i\theta}$ with $0\leq \theta\lt 2\pi$. Then vector addition is really: $$e^{i\theta} \oplus e^{i\eta} = \left\{\begin{array}{ll} e^{i(\theta+\eta)} &\text{if }\theta+\eta\lt 2\pi,\\ e^{i(\theta+\eta-2\pi)}&\text{if }\theta+\eta\geq 2\pi. \end{array}\right.$$ and the scalar product it given by $$r\odot e^{i\theta} = e^{i(r\theta - 2k\pi)}$$ where $k$ is an integer such that $0\leq r\theta - 2k\pi \lt 2\pi$.
Now, take $e^{i\pi}$ and consider multiplication by $2$ and then by $\frac{1}{2}$. We get $$\frac{1}{2}\odot(2\odot e^{i\pi}) = \frac{1}{2}\odot (e^{0i}) = e^{0i}.$$ However, $$\left(\frac{1}{2}\cdot 2\right)\odot e^{i\pi} = 1\odot e^{i\pi} = e^{i\pi}.$$ So the scalar multiplication is not really associative on $S^1$ for this vector space structure.