The MO page here addresses your question (answer and comments).
By the way, your characterization of Haar measure is missing regularity conditions. Of course the main technical property of Haar measure that we care about is its invariance under left (or right) multiplication, but you need further properties to define a Haar measure. For example, according to your definition the counting measure on every topological group is a Haar measure! It is a common "error" to describe Haar measure in such a way that counting measure fits the description even though counting measure is not a Haar measure on $G$ when $G$ is not discrete.
What you left out are the following two conditions on $\mu$:
(i) for each Borel subset $A \subset G$, $\mu(A) = \inf \mu(U)$ where $U$ runs over open subsets of $G$ containing $A$,
(ii) for each open or $\sigma$-finite subset $A \subset G$, $\mu(A) = \sup \mu(K)$ where $K$ runs over compact subsets of $G$ contained in $A$.
These conditions taken together are called $\sigma$-regularity, and a Haar measure on $G$ is a $\sigma$-regular Borel measure on $G$ fitting the conditions you gave. Such conditions should be found in a proof that Haar measure is unique up to a scaling factor. Something like these conditions had better be used in a proof of uniqueness, because without them counting measure would be a Haar measure for $\mathbf R$ with its usual topology, which is false.
A measure that satisfies (i) and (ii) for all Borel subsets $A \subset G$ is called regular. Regularity is a stronger condition than $\sigma$-regularity and Haar measure on some locally compact groups is not regular. For example, let $G = S^1 \times \mathbf R_d$ where $\mathbf R_d$ is $\mathbf R$ with the discrete topology. Topologically this is like a cylinder of "independent circles".
Counting measure is a Haar measure on $\mathbf R_d$ and on $S^1$ there is a standard normalized Haar measure $d\theta/2\pi$. The product of these is a Haar measure on $G$ and it is not regular: the subset $A = \{1\} \times [0,1]$ of $G$ is closed (and not open), so $A$ is a Borel subset of $G$. Check $\mu(K) = 0$ for all compact $K \subset A$ and $\mu(U) = \infty$ for all open $U \supset A$. Thus condition (ii) for a Haar measure does not apply to $A$ (which is neither open nor $\sigma$-finite, so there is no contradiction).
Best Answer
Note that for any Borel set $A \subseteq \mathbb{R}^*$ we can define $$\mu (A) = \int_A \frac{1}{|x|} dx$$ where $dx$ stands for integration against the Lebesgue measure. This measure is Borel regular and locally finite; as advertised, you can argue by change of variables to show that this measure is translation invariant.
You may conclude that this is indeed a Haar measure. Notice that since $\mathbb{R}^*$ is an Abelian group this measure is, at once, the left and right Haar measure for this group.