Yes it is bounded. Assume throughout that $\lambda<1$ and split
$$J(\lambda)=\int_0 ^1 \sin(y) \log(y) g(\lambda y)dy + \int_1 ^\infty \sin(y) \log(y) g(\lambda y)dy=:J_0(\lambda)+J_1(\lambda) .$$
In $[0,1]$ we have that $0<\lambda y <\lambda <1$ so that $|g(\lambda y)|\lesssim_g 1$ and this is best possible since $g(0)\neq 0$. We estimate $J_0$ by brute force
$$|J_0(\lambda)|\lesssim_g \int_0 ^1 |\sin(y)\log(y)|dy\leq\int_0 ^1 y \log(1/y)dy\lesssim 1.$$
For $J_1$ we first integrate by parts once using that $\lim_{y\to\infty}g(y)=0$:
$$J_1(\lambda)=-\cos(y)\log(y)g(\lambda y)\bigg|_{1} ^\infty +\int_1 ^\infty \cos(y)\frac{d}{dy}\big(\log (y) g(\lambda y)\big)dy$$
$$ = \int_1 ^\infty \cos(y)\frac{g(\lambda y)}{y}dy +\lambda \int_1 ^\infty \cos(y)\log(y) g'(\lambda y)dy=:A+B.$$
For $A,B$ it is convenient to split the interval of integration to $[1,1/\lambda]$ and $[1/\lambda,\infty]$ where different bounds for $g(\lambda \cdot)$ apply and integrate by parts as many times as necessary.
$$A_1:=\int_1 ^\frac{1}{\lambda} \cos(y)\frac{g(\lambda y)}{y}dy= \sin(y) \frac{g(\lambda y)}{y}\bigg|_{1} ^{1/\lambda}-\int_1 ^\frac{1}{\lambda} \sin(y) \big(\frac{\lambda g'(\lambda y)}{y}-\frac{g(\lambda y)}{y^2} \big)dy$$
$$=\lambda\sin(1/\lambda)g(1)-\sin(1)g(\lambda)-\lambda\int_1 ^\frac{1}{\lambda}\sin(y)\frac{g'(\lambda y)}{y}dy +\int_1 ^\frac{1}{\lambda}\sin(y)\frac{g(\lambda y)}{y^2}dy.$$
Since $|\lambda y|< 1$ and $g$ is real analytic we have $|g'(\lambda y)|+|g(\lambda y)|\lesssim_g 1$ for $y\in[1,1/\lambda]$. Using these estimates we get $|A_1|\lesssim_g 1 $ (one can be more careful here in order to get the main term).

Similarly
$$A_2:=\sin(y)\frac{g(\lambda y)}{y}\bigg|_{\frac{1}{\lambda}}^\infty - \int _ {\frac{1}{\lambda}} ^\infty \sin(y)\frac{d}{dy}\big( \frac{g(\lambda y)}{y}\big)dy $$
so that $|A_2|\lesssim \lambda |g(1)|\lesssim_g \lambda.$ Now for $B$ we split again
$$B_1:= \lambda \sin(y) \log(y) g'(\lambda y) \bigg|_{1} ^{1/\lambda}-\lambda\int_1 ^\frac{1}{\lambda}\sin(y)\frac{d}{dy}\big(\log(y)g'(\lambda y)\big)dy$$
which shows that $|B_1|\lesssim \lambda \sin(1/\lambda)\log(1/\lambda)|g'(1)|\lesssim_g \lambda \log(1/\lambda).$

Finally
$$B_2:= \lambda \sin(y)\log(y)g'(\lambda y)\big|_\frac{1}{\lambda} ^\infty-\lambda \int _\ frac{1}{\lambda} ^\infty \sin(y)\bigg(\frac{g'(\lambda y)}{y}+\log(y)\lambda g''(\lambda y)\bigg)dy.$$

Using the asmyptotic decay for the derivatives of $g(y)$, $g^{(n)}(x)=O(x^{-2-n})$ for large $y$, and the boundedness for small $y$ we can estimate
$$|B_2|\leq |g'(1)|\lambda \log(1/\lambda)+\lambda \int_{\frac{1}{\lambda}} ^\infty \frac{1}{\lambda^3 y^4}dy+\lambda^2 \int_{\frac{1}{\lambda}} ^\infty \frac{\log(y)}{\lambda^4 y^4}dy $$
$$\lesssim_g \lambda \log(1/\lambda)+\lambda+\lambda\log(1/\lambda).$$
So $|A|\lesssim_g 1$ and $|B|\lesssim_g \lambda \log(1/\lambda)$ as $\lambda \to 0^+$ which implies that $|J(\lambda)|\lesssim_g 1$.

If the function $g$ is something like $g(x)=1/(1+x^2)$ (which satisfies all the hypotheses), I am pretty sure that you also have $|J(\lambda)|\gtrsim 1$ but I don't have the patience to check the details.