Write
\begin{align*}
S_\beta &=(1-\beta)\big[X_0+X_1+X_2+\dots\big]
\\
&=(1-\beta)\big[(1-\beta)X_0 + (\beta-\beta^2)(X_0+X_1) + (\beta^2-\beta^3)(X_0+X_1+X_2) + \dots\big]
\\
&=(1-\beta)^2\big[X_0 + \beta(X_0+X_1) + \beta^2 (X_0+X_1+X_2)+ \dots\big].
\end{align*}

Suppose $X_n$ has finite mean $\mu$. Let $\epsilon>0$. By the Strong Law of Large Numbers, with probability $1$, $X_0+X_2+\dots+X_{n-1}>n(\mu-\epsilon)$ for all large enough $n$. In that case,
\begin{align*}
\liminf_{\beta\uparrow 1} S_\beta
&\geq \liminf_{\beta\uparrow 1} (\mu-\epsilon)(1-\beta)^2[1+2\beta+3\beta^2+\dots]\\
&=\mu-\epsilon.
\end{align*}
In a similar way, with probability $1$, $\liminf_{\beta\uparrow1} S_\beta\leq \mu+\epsilon$. So you have that with probability $1$, the limit $\lim_{\beta\uparrow1} S_\beta$ exists and equals $\mu$.

In the case where the mean is infinite, the same argument gives that with probability $1$, $S_{\beta} \to \infty$ as $\beta\uparrow1$. This seems to contradict your statement that the limit is finite whenever $\mathbb{E} \log_+ X_n<\infty$.