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). Landau's proof is similar to the proof of the Prime Number Theorem.

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).

It is not difficult to adapt Landau's proof and obtain $$(\star)\, \sum_{n \le x} \beta(n) = \frac{1}{\pi} \int_{1/2}^{1} \frac{\sqrt{L(\sigma,\chi_{-4})} \sqrt{\zeta(\sigma)(\sigma-1)} G(\sigma)}{(1-\sigma)^{1/2}}\frac{x^{\sigma}}{\sigma}d\sigma + O\left( \frac{x}{L(x)^c}\right)$$ where $L(x)$ is as in your question. In other words, the statement you quote holds with $$\lambda(t) := \frac{1}{\pi}\sqrt{L(1-t,\chi_{-4})} \sqrt{\zeta(1-t) t} G(1-t) \frac{1}{1-t}.$$ Two references for $(\star)$ (or very similar statements): 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. $(\star)$ has first appeared implicitly in Ramachandra's paper. See Remark 2 in Ramachandra for a discussion of an asymptotic expansion for your integral, as well as the rest of this answer.

The error term in $(\star)$ ultimately comes from the zero-free region for $\zeta(s)L(s,\chi_{-4})$, and in particular the error term is much smaller under GRH.

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 $\log x$.

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 $$K = \frac{1}{\sqrt{2}}\prod_{k\ge 1} \left( \frac{(1-2^{-2^k })\zeta(2^k)}{L(2^k,\chi_{-4})}\right)^{2^{-k-1}}.$$