This result may have been known to Landau. In 1908, Landau proved that $$\sum_{n \le x} \beta(n) \sim K \frac{x}{\sqrt{\log x}}$$ holds as $x \to \infty$, where $K$ is the Landau-Ramanujan constant given by $$K = \frac{1}{\sqrt{2}} \prod_{p \equiv 3 \bmod 4}(1-\frac{1}{p^2})^{-\frac{1}{2}}.$$ It is named so because in Ramanujan's first letter to Hardy in 1913, he stated the same result (without proof).
Let $D_{\beta}(s):=\sum_{n=1}^{\infty} \beta(n)/n^s$. Recall that $$D_{\beta}(s) = (1-2^{-s})^{-1}\prod_{p \equiv 1 \bmod 4} (1-p^{-s})^{-1} \prod_{p \equiv 3 \bmod 4} (1-p^{-2s})^{-1}$$ holds for $\Re s > 1$. Hence $$D_{\beta}(s) = \sqrt{\zeta(s) L(s,\chi_{-4})} G(s)\quad \text{ for }\quad G(s) := \sqrt{\frac{1}{1-2^{-s}}} \prod_{p \equiv 3 \bmod 4} (1-p^{-2s})^{-\frac{1}{2}}.$$ It is not hard to see that $G(s)$ converges absolutely for $\Re s \ge 1/2+\varepsilon$ and defines a bounded, analytic function there. Moreover, $\sqrt{L(s,\chi_{-4})}$ has an analytic continuation within any zero-free region, and the same goes for $\sqrt{\zeta(s)(s-1)}$ (since $L(s,\chi_{-4})$ and $\zeta(s)(s-1)$ are entire).
While I haven't checked Landau's paper recently, I would expect it is not difficult to extract from his proof that the following holds: $$(\star)\, \sum_{n \le x} \beta(n) = \frac{1}{\pi} \int_{1/2}^{1} \frac{\sqrt{L(\sigma,\chi_{-4})} \sqrt{\zeta(\sigma)(\sigma-1)} G(\sigma)}{(1-\sigma)^{1/2}}\frac{x^{\sigma}}{\sigma}d\sigma + O\left( \frac{x}{L(x)^c}\right)$$ where $L(x)$ is as in your question. In other words, the statement you quote holds with $$\lambda(t) := \frac{1}{\pi}\sqrt{L(1-t,\chi_{-4})} \sqrt{\zeta(1-t) t} G(1-t) \frac{1}{1-t}.$$ Three references for $(\star)$ (or very similar statements): Ramachandra's paper, Exercise 21(d) on p. 187 of Montgomery and Vaughan's book "Multiplicative Number Theory I" and Theorem 2.1 of David-Devin-Nam-Schlitt. See Remark 2 in Ramachandra for a discussion of an asymptotic expansion for your integral, as well as the rest of this answer.
Landau's proof is a special case of the Selberg-Delange method, which is why some authors call this method the Landau-Selberg-Delange method.
The error term in $(\star)$ ultimately comes from the zero-free region for $\zeta(s)L(s,\chi_{-4})$. (So strictly speaking, because Landau only had the classical zero-free region, he would only get an error of $O(x \exp(-c\sqrt{\log x}))$.) The error term is much smaller under GRH, see Appendix B2 here.
Observe that $\lambda(0) = G(1)\sqrt{\pi/4}/\pi= K/\sqrt{\pi}$. This is consistent with Landau's result, since $$\int_{0}^{1/2}x^{1-t} \frac{\lambda(t)}{\sqrt{t}} dt = \frac{x}{\sqrt{\log x}}\int_{0}^{(\log x)/2}e^{-v} \frac{\lambda(v/\log x)}{\sqrt{v}}dv \sim \frac{x}{\sqrt{\log x}} \lambda(0) \int_{0}^{\infty}e^{-v} \frac{dv}{\sqrt{v}},$$ and the last integral is $\Gamma(1/2)=\sqrt{\pi}$. By Taylor-expanding $\lambda$ at $0$, one can obtain an asymptotic expansion for $\sum_{n\le x}\beta(n)$ in descending powers of $\log x$.
To address Steven Clark's question from the comments: Shanks, and independently Flajolet and Vardi, proved that $$D_{\beta}(s) = \sqrt{\zeta(s)L(s,\chi_{-4})(1-2^{-s})^{-1}}\prod_{k\ge 1} \left( \frac{(1-2^{-2^k s})\zeta(2^ks)}{L(2^k s,\chi_{-4})}\right)^{2^{-k-1}}$$ holds for $\Re s >1$ from which they deduced that $$K = \frac{1}{\sqrt{2}}\prod_{k\ge 1} \left( \frac{(1-2^{-2^k })\zeta(2^k)}{L(2^k,\chi_{-4})}\right)^{2^{-k-1}}.$$