This is basically Second construction 5.3 from Eric K. van Douwen. *Finitely additive measures on $\mathbb N$*. Topology Appl., **47 (3)**, (1992), 223–268. MR1192311 (94c:28004). This issue of Topology and Its Applications was a special issue dedicated to Eric K. van Douwen.

**EDIT:** Only after making the construction I posted below I returned to the paper and noticed that it contains Appendix 5C: The number of elastic measures. It is shown here that the set of elastic measures has cardinality $2^{\mathfrak c}$. Every elastic measure has properties required in the above question.

So this gives a better result than I obtained below. I have decided to keep the original version of my answer here. Just in case some of the information given there might be interesting for some people reading this question.

$\newcommand{\Flim}{\operatorname{\mathcal F-\lim}}\newcommand{\FF}{\mathcal F}$
The author uses diffuse mean in this construction. I will use $\FF$-limit, which is also a diffuse mean. (See that paper for terminology.) I will summarize which properties of limit along an ultrafilter I need. (Of course, you can use some other functional fulfilling these properties instead of $x\mapsto \Flim x_n$. I believe that such functionals can be obtained, for example, using Hahn-Banach theorem.)

For any filter $\FF$ we can define the $\FF$-limit of a sequence $(x_n)$ as
$$\Flim x_n=L \qquad\Leftrightarrow\qquad (\forall \varepsilon>0) \{n; |x_n-L|<\varepsilon\}\in\FF.$$
If $\FF$ is a *free ultrafilter*, then $\Flim x_n$ exists for each bounded sequence. Moreover, we know that for a free ultrafilter $\FF$ we have

- If $(x_n)$ is convergent, then $\Flim x_n=\lim x_n$ (=extends limit);
- $\Flim (x_n+y_n) = \Flim x_n + \Flim y_n$ and $\Flim (cx_n)=c\Flim x_n$ (=linearity);
- $x_n\le y_n$ $\Rightarrow$ $\Flim x_n\le\Flim y_n$ (=is positive);
- For a given sequence $(x_n)$ and any cluster point $c$ of $(x_n)$ there exists a free ultrafilter such that $\Flim x_n=c$.

Some references for $\FF$-limits can be found here or in the comments to this post. (There is also a Wikipedia article on ultralimit. However, this article discusses ultralimits of sequences only very briefly.)

For $A\subseteq\mathbb N$ we define
$$\mu_n(A) = \frac{\sum_{j\le n}\frac1j \chi_A(j)}{\ln n}.$$

Notice that $\overline\delta(A)=\limsup\limits_{n\to\infty} \mu_n(A)$ is precisely the upper logarithmic density. For limit inferior we get the lower logarithmic density $\underline\delta(A)=\liminf\limits_{n\to\infty} \mu_n(A)$.

Let $\FF$ be a free ultrafilter. We define
$$\mu(A)=\Flim \mu_n(A).$$
(Maybe $\mu_F$ would be a better notation, since it depends on the choice of $\FF$, but I will use $\mu$ for brevity.)

It is relatively easy to see that $\mu(A)$ is a finitely additive measure on $\mathbb N$.

We will show below that $\mu(kA+h)=\frac1k\mu(A)$.

Now if we fix some set $A$ such that $\underline\delta(A)=0$ and $\overline\delta(A)=1$, then every point in the interval $[0,1]$
is a cluster point of the sequence $(\mu_n(A))$. (Note that the set of all cluster points of this sequence is connected.)

So there are at least $\mathfrak c$ such measures, since $\mu(A)$ can attain all values between $0$ and $1$ (for various choices of the free ultrafilter $\FF$).

So it remains to show that the measures of the form described above fulfill the condition
$$\mu(kA+h)=\frac1k\mu(A).$$
In fact, this is shown van Douwen's paper. But since I changed notation a little bit I will include sketch of the proof.

**$\mu$ is shift-invariant, i.e., $\mu(A+1)=\mu(A)$.**

Let $B=A+1$. We have
$$|\mu_n(A)-\mu_n(B)| \le \frac{\frac1n+\sum_{k<n} \left(\frac1k-\frac1{k+1}\right)}{\ln n}\le\frac1{\ln n}.$$
This implies that $\lim\limits_{n\to\infty} (\mu_n(A)-\mu_n(B)) =0$, hence $\Flim (\mu_n(A)-\mu_n(B))=0$ which means that $\Flim\mu_n(A)=\Flim\mu_n(B)$ and $$\mu(B)=\mu(A).$$

**$\mu$ is $\mathbb N$-scale invariant, i.e., $\mu(kA)=\frac1k\mu(A)$ for $k\in\mathbb N$.**

Let $B=kA$. Now we have $$\mu_n(B)=\frac{\sum_{j\le n}\frac1j \chi_B(j)}{\ln n} =
\frac{\sum_{j\le n/k}\frac1{kj} \chi_A(j)}{\ln n} = \frac1k \frac{\sum_{j\le n/k}\frac1j \chi_A(j)}{\ln n}.$$
This yields
$$|\frac1k\mu_n(A)-\mu_n(B)| \le \frac1k \frac{\sum_{n/k<j\le n} \frac1j}{\ln n} \sim \frac1k \cdot \frac{\ln k}{\ln n}.$$
Again $\lim\limits_{n\to\infty} (\frac1k\mu_n(A)-\mu_n(B)) =0$ and $\mu(B)=\frac1k\mu(A)$.