If you wish to avoid delta functions, you could just Fourier transform, $f(k)=\int_{-\infty}^\infty e^{ikx}y(x)dx$. The differential equation then transforms into

$$(k^2+\lambda^2) f(k)=y(0)$$

Divide both sides of the equation by $k^2+\lambda^2$, integrate over $k$ and you arrive at

$$\int_{-\infty}^\infty f(k)dk=2\pi y(0) = \frac{\pi}{|\lambda|}y(0)$$

Since $y(0)\neq 0$ (otherwise the solution is not normalizable), you can conclude that $\lambda=\pm 1/2$.

If you wish to avoid delta functions, you could just Fourier transform, $f(k)=\int_{-\infty}^\infty e^{ikx}y(x)dx$. The differential equation then transforms into

$$(k^2+\lambda^2) f(k)=y(0)$$

Divide both sides of the equation by $k^2+\lambda^2$, integrate over $k$ and you arrive at

$$\int_{-\infty}^\infty f(k)dk = y(0)\int_{-\infty}^\infty \frac{1}{k^2+\lambda^2}dk$$ $$\Rightarrow 2\pi y(0)=\frac{\pi}{|\lambda|}y(0)$$

Since $y(0)\neq 0$ (otherwise $y(x)=0$ for all $x$), you can conclude that $\lambda=\pm 1/2$.