Haar Measure on Locally Compact Semigroups

Not every locally compact semigroup admits a (locally finite) left-invariant measure. In fact, this has nothing to do with any sort of analytic technicalities and already fails for finite semigroups. For example, consider $S=\{a,b\}$ with $ab=a^2=a$ and $ba=b^2=b$. Then $aS=\{a\}$ and $bS=\{b\}$, so no (finite) measure on $S$ can have $\mu(S)=\mu(aS)=\mu(bS)$.

If memory serves a locally compact semigroup admits a left invariant Borel measure iff it has a minimal ideal which is a direct product of a group and a right zero semigroup.

The measure is supported on the minimal ideal.

Update:

The result I was trying to recall is from http://www.ams.org/journals/proc/1966-017-02/S0002-9939-1966-0188341-7/S0002-9939-1966-0188341-7.pdf.

A locally finite Borel measure $\mu$ on a locally compact inverse semigroup $S$ is said to be left invariant if $\mu(s^{-1}B)=\mu(B)$ for all Borel sets $B$ where $s^{-1}B=\{x\in S\mid sx\in B\}$.

It is proved in the above paper that if $S$ is locally compact and left translations are proper (i.e. $s^{-1}K$ is compact when $K$ is compact), then $S$ has a left invariant measure iff it has a unique minimal left ideal. This is the same as saying it has a minimal ideal which is a direct product of a group and a right zero semigroup. You can take as the measure the product of Haar measure on the semigroup with your favorite locally finite Borel measure on the right zero semigroup.