Here it is another proof: it works when $\mu$ and $\omega$ are finite measures.

It is straightforward to see that 
$$ E(\mu | \omega ) + \mu (M) = \sup \left\{ \int_M f d \mu - \int_M e^{f} d \omega : f \in L^{\infty} ( \omega + \mu ) \right\}. $$

This is rather easy to prove: it relies on the fact that for $a \geq0$, we have $at-e^t \leq a \log (a) - a$ and the maximum is obtained when $t=log(a)$. Moreover, by a density argument we can prove that

$$ E(\mu | \omega ) + \mu (M) = \sup \left\{ \int_M f d \mu - \int_M e^{f} d \omega : f \in C_b(M) \right\}. $$

Since the right hand side is the supremum of continuous functions in both $\mu$ and $\nu$ we can deduce that if $\mu_n \rightharpoonup \mu$ and $\omega_n \rightharpoonup \omega$ then 
$$ \liminf_n E(\mu_n | \omega_n) \geq E(\mu | \omega),$$
that is, the relative entropy is jointly semicontinuous. Moreover we expressed the entropy as a supremum of linear functions in $(\mu, \omega)$ and so we have that it is convex in the couple $(\mu, \omega)$, that is
$$E(t\mu + (1-t)\mu' | t\omega + (1-t)\omega' ) \leq tE(\mu|\omega) + (1-t) E(\mu'|\omega');$$
In particular it is convex in the first variable