Your sequence is
$$x[n]=a^n\sin(n\omega_0)u[n]\tag{1}$$
First of all, note that the Discrete-Time Fourier Transform (DTFT) of $(1)$ only exists if $|a|<1$. (The case $|a|=1$ can be handled by using Delta impulses). Anyway, for $|a|>1$ the DTFT of $(1)$ does not exist. Second, you can write $x[n]$ as the multiplication of $x_1[n]=a^nu[n]$ and $x_2[n]=\sin(n\omega_0)$. The DTFTs of these two sequences are given by
$$\begin{align}X_1(e^{j\omega})&=\frac{1}{1-ae^{-j\omega}},\quad |a|<1\\
X_2(e^{j\omega})&=\frac{\pi}{j}\left[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)\right],\quad -\pi <\omega <\pi\end{align}\tag{2}$$
Of course, $X_2(e^{j\omega})$ is also $2\pi$-periodic, but in $(2)$ only the interval $-\pi<\omega<\pi$ is considered (assuming $0<\omega_0<\pi$).
Note that in your expression for $X_2(e^{j\omega})$ you have a sign error. Furthermore, note that $X_2(e^{j\omega})$ is the DTFT of $x_2[n]=\sin(n\omega_0)$, and not of $x_2[n]=\sin(n\omega_0)u[n]$, as claimed in your question.
Now we have
$$X(e^{j\omega})=\frac{1}{2\pi}X_1(e^{j\omega})\star X_2(e^{j\omega})=\frac{1}{2\pi}\int_{-\pi}^{\pi}X_1(e^{j\theta})X_2(e^{j(\omega-\theta)})d\theta\tag{3}$$
The only thing you need to know to compute $(3)$ is the property
$$F(\omega)\star \delta(\omega-\omega_0)=F(\omega-\omega_0)\tag{4}$$
for any function $F(\omega)$. With $(4)$, evaluating $(3)$ with the DTFTs in $(2)$ results in
$$\begin{align}X(e^{j\omega})&=\frac{1}{2\pi}\frac{\pi}{j}\left[X_1\left(e^{j(\omega-\omega_0)}\right)-X_1\left(e^{j(\omega+\omega_0)}\right)\right]\\&=\frac{1}{2j}\left[\frac{1}{1-ae^{-j(\omega-\omega_0)}}-\frac{1}{1-ae^{-j(\omega+\omega_0)}}\right]\tag{5}\end{align}$$
The two terms in $(5)$ can be combined resulting in
$$X(e^{j\omega})=\frac{ae^{-j\omega}\sin(\omega_0)}{1-2a\cos(\omega_0)e^{-j\omega}+a^2e^{-2j\omega}}\tag{6}$$
