I don't know whether this helps or whether you've already seen this before, but this made a lot more intuitive sense to me than the combinatorial approach in Halmos's book.

The key point in the proof is to prove the maximal ergodic theorem.  This states that if $M_T$ is the maximal operator $M_T f= \sup_{n >0} \frac{1}{n+1} (f + Tf + \dots + T^n f)$, then $\int_{M_T f>0} f \geq 0$.  Here $T$ is the associated map on functions coming from the measure-preserving transformation.

This is a weak-type inequality, and the fact that one considers the maximal operator isn't terribly surprising given how they arise in a) the proof of the Lebesgue differentation theorem (namely, via the Hardy-Littlewood maximal operator $Mf(x) = \sup_{t >0} \frac{1}{2t} \int_{x-t}^{x+t} f(r) dr$.  In the theory of singular integrals, one can define a maximal operator in the same way and prove that it is $L^p$-bounded for $1 < p < \infty$ and weak-$L^1$ bounded in suitably nice homogeneous cases (e.g. the Hilbert transform).  One of the consequences of this is, for instance, that the Hilbert transform can be computed a.e. via the Cauchy principal value of the usual integral.  I'm pretty sure the boundedness of the maximal operator of the partial sums of Fourier series is used in the proof of the Carleson-Hunt theorem.  So using maximal operators (and, in particular, weak bounds on them) to establish convergence is fairly standard.  Once the maximal inequality has been established, it isn't usually very hard to get the pointwise convergence result, and the ergodic theorem is no exception.

The maximal ergodic theorem actually generalizes to the case where $T$ is an operator of $L^1$-norm at most 1, and thinking of it in a more general sense might meet the criteria of your question. In particular, let $T$ be as just mentioned, and consider $M_T$ described in the analogous way. Or rather, consider $M_T'f = \sup_{n \geq 0} \sum_{i=0}^n T^if$. Clearly $M_T'f >0$ iff $M_Tf >0$.  Moreover, $M_T'$ has the crucial property that $T M_T' f + f = M_T' f$ whenever $M_Tf>0$.  

Therefore, 
$\int_{M_T'f>0} f = \int_{M_T'f>0} M_T'f - \int_{M_T'f>0} TM_T'f.$  The first part is in fact $||M_T'f||_1$ because the modified maximal operator is always nonnegative.  The second part is at most $||T M_T'f|| \leq ||M_T'f||$ by the norm condition. Hence the difference is nonnegative.


Perhaps this will be useful: let $M$ be an operator (not necessarily linear) sending functions to nonnegative functions such that $(T-I)Mf  = f$ wherever $Mf>0$, for $T$ an operator of $L^1$-norm at most 1).  Then $\int_{Mf>0} f \geq 0$.  The proof is the same.