Let $\mu_n$ be the measure defined by $$\mu_n(A) = \frac{1}{\tau_n} \int_0^{\tau_n} \mu(A-t)\,dt.$$
In general, this sequence doesn't converge weakly. Suppose for instance that $\mu$ is a point mass at $0$. Then $\mu_n$ is the uniform probability measure on $[0, \tau_n]$. Let $f$ be any bounded, continuous, strictly positive, Lebesgue integrable function on $\mathbb{R}$ (for instance, $f(x) = e^{-x^2}$ would do). Then $$\int_\mathbb{R} f\,d\mu_n = \frac{1}{\tau_n} \int_0^{\tau_n} f(t)\,dt \le \frac{1}{\tau_n} \int_{\mathbb{R}} |f(t)|\,dt \to 0$$ as $\tau_n \to \infty$. Since each $\mu_n$ is a positive measure, if there is a weak limit $\mu^*$ it is also a positive measure. So this would force $\int f\,d\mu^* = 0$, which can only happen if $\mu^* = 0$.
But $\mu_n$ doesn't converge to 0 either; take $f=1$. So the sequence doesn't converge weakly at all.
If you had $\tau_n \to 0$ (which is how I originally misread the question) then $\mu_n \to \mu$ weakly:
Verify that $\mu_n$ is indeed a measure (i.e. that it is countably additive). Use the monotone convergence theorem.
For a bounded measurable function $f : \mathbb{R} \to \mathbb{R}$, verify that $$\int_{\mathbb{R}} f\,d\mu_n = \frac{1}{\tau_n} \int_0^{\tau_n}\int_{\mathbb{R}} f(x+t)\,\mu(dx)\,dt.$$ Start with the case where $f=1_A$ is an indicator function, then do simple functions, then nonnegative measurable functions, then bounded measurable functions (the "standard mantra").
Use Fubini's theorem to verify that $$\tag{*} \int_{\mathbb{R}} f\,d\mu_n = \int_{\mathbb{R}} \frac{1}{\tau_n} \int_0^{\tau_n}f(x+t)\,dt\,\mu(dx).$$
For bounded continuous $f$, set $f_n(x) = \frac{1}{\tau_n} \int_0^{\tau_n}f(x+t)\,dt$. Use the continuity of $f$ to show that $f_n \to f$ pointwise, and the sequence is bounded by $\|f\|_\infty$.
Use the dominated convergence theorem on (*) to show that $\int_\mathbb{R} f\,d\mu_n = \int_{\mathbb{R}} f_n\,d\mu \to \int_\mathbb{R} f\,d\mu$. Hence $\mu_n \to \mu$ weakly.