Clearly, the expected time $w_n$ till extinction from the initial state $n$ is nondecreasing in $n$. So, if $w_1=\infty$, then $w_n=\infty$ for all natural $n$, so that the difference $w_{n+1}-w_n$ makes no sense, and hence 
the desired conclusion 
\begin{equation}
\lim_{n\to\infty}\prod_{j=1}^n\frac{\lambda_j}{\mu_j}(w_n-w_{n+1)}=0 \tag{1}	
\end{equation}
makes no sense either. 

It remains to consider the case $w_1<\infty$. Then the equation 
\begin{equation}
	w_n=\frac{\lambda_nw_{n+1}+\mu_nw_{n-1}+1}{\lambda_n+\mu_n} \tag{2}
\end{equation}
(together with the condition $w_0=0$) 
implies that $w_n<\infty$ for all natural $n$. 
For natural $j$, let then 
$$h_j:=w_j-w_{j-1},\quad\pi_j:=\prod_{i=1}^{j-1}\frac{\lambda_i}{\mu_i}  
$$
(so that $\pi_1=1$), and 
$$u_j:=\pi_j h_j.$$
Then (2) can be rewritten as 
$$u_{n+1}=u_n-\frac{\pi_n}{\mu_n}, 
$$
which implies
\begin{equation}
u_n=u_1-\sum_{j=1}^{n-1}\frac{\pi_j}{\mu_j}. \tag{3}	
\end{equation}


Also, the equality 
$$w_1=\frac{1}{\mu_1}\sum_{n=0}^\infty\prod_{j=1}^n\frac{\lambda_j}{\mu_{j+1}}$$
can be rewritten as 
$$u_1[=h_1=w_1]=\sum_{j=1}^\infty\frac{\pi_j}{\mu_j}.  
$$
This and (3) imply
$$\pi_n h_n=u_n=\sum_{j=n}^\infty\frac{\pi_j}{\mu_j}. 
$$
Now the case condition $w_1<\infty$ implies $\pi_n h_n\underset{n\to\infty}\longrightarrow0$, which means that the desired conclusion (1) holds.