I will elaborate on this. Following Ma's suggestion, from the Cauchy-Schwarz inequality we have
\begin{align}
\bigg|\int_{t}^{t+\delta}{e^{- \lambda(t+\delta-s)}x(s)ds}\bigg| &\leq \int_{t}^{t+\delta}{\big|e^{- \lambda(t+\delta-s)}x(s)\big|ds}\\
&\leq \sqrt{\int_{t}^{t+\delta}{e^{-2\lambda (t+\delta-s)}ds}\int_{t}^{t+\delta}{x^2(s)ds}}\\
&= \sqrt{\frac{1-e^{-2\lambda \delta}}{2\lambda}}\sqrt{\int_{t}^{t+\delta}{x^2(s)ds}}.
\end{align}
For a square integrable $x$ it holds true that $\lim_{t\rightarrow\infty}\int_t^{t+\delta}{x^2(s)ds}=0$.
Using this fact and taking the limit $t\rightarrow\infty$ in the inequality above we obtain $\lim_{t\rightarrow\infty}\int_{t}^{t+\delta}{e^{- \lambda(t+\delta-s)}x(s)ds}=0$.
We can write $\int_0^{t+\delta}{e^{-\lambda(t+\delta-s)}x(s)ds}=e^{-\lambda\delta}\int_0^t{e^{-\lambda (t-s)}x(s)ds}+\int_{t}^{t+\delta}{e^{- \lambda(t+\delta-s)}x(s)ds}$.
If we now define $S:=\lim_{t\rightarrow\infty}\int_0^t{e^{-\lambda (t-s)}x(s)ds}$ and take the limit in the identity above we have $S=e^{-\lambda\delta}S$ that yields the desired $S=0$. However, one has to prove that the limit $S$ actually exists.