What is the relationship between Haar Measure and the Birkhoff-Kakutani Theorem?

Since your group is locally compact (and Hausdorff!), the given metric $d$ on $G$ always has the property that for some $r>0$ the closed ball $\bar{B}(e,r)\subset G$ is compact. This does not imply that the same holds for $r=1$. For instance, take an infinite discrete group equipped with discrete metric. If you really like $r=1$, you can just rescale $d$ by replacing it with $r^{-1}d$. As for your last question about bounded diameter for finite measure sets: This is a very unnatural requirement. Just think of the Lebesgue measure on ${\mathbb R}^2$. As you know from calculus, there are many unbounded subsets of finite area in the plane. Of course, you can replace the Euclidean metric with an equivalent invariant bounded metric so that every subset has bounded diameter, but why would you want to do this?

There are other ways in which metric and measure can be related as in "metric measure spaces"; the doubling property will be sometimes true (e.g. for Lie groups), but sometimes it will fail, e.g. for some infinite-dimensional compact groups (yes, there are such things!).

Struble's theorem: on a locally compact group with countable basis, there always exists a left-invariant proper compatible metric.

Compatible means: defining the topology. Proper means: whose closed balls are compact (and thus of finite Haar measure). See Theorem 2.B.4 in my book with Pierre de la Harpe (arXiv link).

(As mentioned by other people, when $G$ is not compact, other compatible metric can fail to be proper, e.g., can be bounded.)