Let $X(t)$ be a Poisson process with rate $x$, $Y(t)$ a Poisson process with rate $y$, and $Z(t)=X(t)+Y(t)$. Then $Z(t)$ is a Poisson process with rate $x+y$. Let $T_n$ be the occurences of $Z(t)$ and $S_0=0$, $S_n = \sum_{i=1}^n T_n$ the associated renewal sequence. The residual life is defined as the interval from $t$ until the next occurence, i.e. $R(t) = S_{N(t)+1}-t$, where $N(t) = \sum_{n=1}^\infty\mathsf 1_{(0,t]}(S_n)$. For a given sample function $r(t)$ of the renewal process, the sample function of residual life decreases linearly with a slope of $-1$ from the beginning to the end of each inter-renewal interval. Hence the integral $\int_0^t y(s)\ \mathsf ds$ is a sum of isoceles right triangles, with part of a final triangle at the end. It follows that
$$
\int_0^t r(s)\ \mathsf d = \frac12\sum_{i=1}^{n(t)}t_i^2 + \int_{s_{n(t)}}^t r(s)\ \mathsf ds,
$$
where $\{t_i\}$ is the set of sample values of the inter-renewal intervals. Since this holds for every sample point, the random variable $\int_0^t R(s)\ \mathsf ds$ can be expressed in terms of the $T_n$ by
$$
\int_0^t R(s)\ \mathsf ds = \frac12\sum_{n=1}^{N(t)}T_n^2 + \int_{S_{N(t)}}^t R(s)\ \mathsf ds.$$
We can rewrite this as
$$
\frac12\sum_{n=1}^{N(t)}T_n^2\leqslant \int_0^t R(s)\ \mathsf ds \leqslant \frac12 \sum_{n=1}^{N(t)+1} T_n^2.
$$
Dividing by $t$ yields
$$
\frac1{2t}\sum_{n=1}^{N(t)}T_n^2\leqslant \frac1t\int_0^t R(s)\ \mathsf ds \leqslant \frac1{2t} \sum_{n=1}^{N(t)+1} T_n^2.
$$
Taking the limit as $t\to\infty$ of the term on the left yields
$$
\lim_{t\to\infty}\frac{\sum_{n=1}^{N(t)}T_n^2}{2t} = \lim_{t\to\infty} \frac{\sum_{n=1}^{N(t)}T_n^2}{2t}\frac{N(t)}{2t}.
$$
Since $N(t)\stackrel{t\to\infty}\longrightarrow 0$ almost surely, the strong law of large numbers implies that
$$
\lim_{t\to\infty} \frac{\sum_{n=1}^{N(t)}T_n^2}{2t}= \lim_{k\to\infty}\frac{\sum_{n=1}^k T_n^2}k = \mathbb E[T^2]\ \mathrm{ a.s.}.\tag1
$$
Now we require a lemma, the so-called strong law for renewal processes. It states that for a renewal process with mean inter-renewal time $\mu<\infty$, $\lim_{t\to\infty}\frac{N(t)}t=\frac1\mu$ a.s. To establish this, we first show that $\lim_{t\to\infty}N(t)=\infty$ a.s. and $\lim_{t\to\infty}\mathbb E[N(t)]=\infty$ a.s. To see this, note that for each $\omega$, $N(t,\omega)$ is a nondecreasing real-valued function of $t$ and thus has a limit (finite or infinite). By the fundamental equivalence $\{S_n>t \}=\{N(t)< n\}$, we have for each positive integer $n$
$$
\lim_{t\to\infty}\mathbb P(N(t)<n) = \lim_{t\to\infty} \mathbb P(S_n>t) = 1-\lim_{t\to\infty}\mathbb P(S_n\leqslant t) = 0.$$ It follows that $$\mathbb P\left(\lim_{t\to\infty}\{N(t(\omega))<n\}\right) = 0, $$
and hence $\lim_{t\to\infty}N(t)=\infty$ a.s. Moreover, $t\mapsto\mathbb E[N(t)]$ is nondecreasing and thus has a (possibly infinite) limit as $t\to\infty$. For each $n$, we may choose $T$ such that $\mathbb P(N(t)\geqslant n)\geqslant\frac12$ for $t\geqslant T$, and hence $\mathbb E[N(t)]\geqslant \frac n2$ for $t\geqslant T$. It follows that $\lim_{t\to\infty}\mathbb E[N(t)] = \infty$. Now for another lemma: if $\{Z_n\}$ is a sequence of random variables that converges almost surely to $\alpha\in\mathbb R$ and $f:\mathbb R\to\mathbb R$ is continuous at $\alpha$, then $$\lim_{n\to\infty}f(Z_n) = f(\alpha)\ \mathrm{a.s.}$$
For if $\{\alpha_n\}$ is a sequence of real numbers that converges to $\alpha$, for every $\varepsilon>0$ there exists $\delta>0$ such that $|z-\alpha|<\delta$ implies $|f(z)-f(\alpha)|<\varepsilon$. Choose $N$ such that $|\alpha_n-\alpha|<\delta$ for $n\geqslant N$. It follows that $\lim_{n\to\infty} f(\alpha_n)=f(\alpha)$. If $\lim_{n\to\infty}Z_n(\omega)=\alpha$, then $\lim_{n\to\infty}f(Z_n(\omega)) = f(\alpha)$. Since the set of such $\omega$ has probability one, the lemma follows.
Now, choose $f(x)=\frac1x$ for $x>0$ - clearly $f$ is continuous at $\mu$. It follows from the preceeding lemma that
$$
\lim_{n\to\infty} \frac n{S_n} = \frac1\mu\ \mathrm{a.s.}.
$$
Since $\lim_{t\to\infty}N(t)=\infty$ a.s., we have
$$
\lim_{t\to\infty}\frac{N(t)}{S_{N(t)}} = \frac1\mu\ \mathrm{a.s.}.
$$
Finally, since $\frac{N(t)}{S_{N(t)}}\leqslant \frac{N(t)}t\leqslant \frac{N(t)}{S_{N(t)+1}}$, from
$$
\lim_{t\to\infty}\frac{N(t)}{S_{N(t)+1}} = \frac1\mu\ \mathrm a.s.
$$
we conclude that $\lim_{t\to\infty}\frac{N(t)}t = \frac1\mu$.
Returning to $(1)$, the strong law for renewal processes yields $\lim_{t\to\infty}\frac{N(t)}{2t} = \frac1{2\mathbb E[T]}$ a.s. Thus
$$
\lim_{t\to\infty} \frac{\sum_{n=1}^{N(t)}T_n^2}{2t} = \frac{\mathbb E[T^2]}{2\mathbb E[T]}\ \mathrm a.s.
$$
Similarly,
$$
\lim_{t\to\infty} \frac{\sum_{n=1}^{N(t)+1}T_n^2}{2t} = \lim_{t\to\infty}\frac{\sum_{n=1}^{N(t)+1}T_n^2}{N(t)+1}\frac{N(t)+1}{N(t)}\frac{N(t)}{2t} = \frac{\mathbb E[T^2]}{2\mathbb E[T]}\ \mathsf{a.s.}
$$
Combining these results, we see that with probability one, the time-average residual life is given by
$$
\lim_{t\to\infty}\frac1t\int_0^t R(s)\ \mathsf ds = \frac{\mathbb E[T^2]}{2\mathbb E[T]}.
$$
Now let $A(t) = t-S_{N(t)}$ be the age of the renewal process at time $t$. By the same analysis as for the residual life, we find that
$$
\lim_{t\to\infty}\frac1t\int_0^t A(s)\ \mathsf ds = \frac{\mathbb E[T^2]}{2\mathbb E[T]}\ \mathrm{a.s.}
$$
Now, to at last prove our result: let $L(t)$ be the duration of the inter-renewal interval containing $t$, i.e. $L(t) = S_{N(t)+1}-S_{N(t)}$. It is clear that $L(t)=A(t)+R(t)$, and thus the time-average of the duration is given by
$$
\lim_{t\to\infty}\frac1t\int_0^t L(s)\ \mathsf ds = \frac{\mathbb E[T^2]}{\mathbb E[T]}\ \mathrm{a.s.}
$$
In our example,
$$
\mathbb E[T^2] = \int_0^\infty t^2(x+y)e^{-(x+y)t}\ \mathsf dt = \frac2{(x+y)^2}
$$
and
$$
\mathbb E[T] = \int_0^\infty t(x+y)e^{-(x+y)t}\ \mathsf dt = \frac1{x+y}.
$$
It follows that
$$
\lim_{t\to\infty}\mathbb E[L(t)] = \frac2{(x+y)^2}\cdot(x+y) = \frac2{x+y},
$$
as desired.
