For natural $j$, let $$h_j:=w_j-w_{j-1},\quad\pi_j:=\prod_{i=1}^{j-1}\frac{\lambda_i}{\mu_i}, $$ so that $\pi_1=1$, and $$u_j:=\pi_j h_j.$$ Then the equation
$$w_n=\frac{\lambda_nw_{n+1}+\mu_nw_{n-1}+1}{\lambda_n+\mu_n}$$ for natural $n$ can be rewritten as $$u_{n+1}=u_n-\frac{\pi_n}{\mu_n}, $$ which implies $$u_n=u_1-\sum_{j=1}^{n-1}\frac{\pi_j}{\mu_j}. \tag{1} $$ 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 (1) imply $$\pi_n h_n=u_n=\sum_{j=n}^\infty\frac{\pi_j}{\mu_j}. $$ So, $\pi_n h_n\underset{n\to\infty}\longrightarrow0$ iff $w_1<\infty$.

Finally, the desired conclusion $$\lim_{n\to\infty}\prod_{j=1}^n\frac{\lambda_j}{\mu_j}(w_n-w_{n+1)}=0$$ can be rewritten as $\lim_{n\to\infty}\pi_n h_n=0$.

Thus, the desired conclusion holds iff $w_1<\infty$.

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.

For natural $j$, let $$h_j:=w_j-w_{j-1},\quad s_j:=\frac{\lambda_j}{\mu_j},\quad\pi_j:=\prod_{i=1}^{j-1}s_i, $$ so that $\pi_1=1$, and $$u_j:=\pi_j h_j.$$ Then the equation
$$w_n=\frac{\lambda_nw_{n+1}+\mu_nw_{n-1}+1}{\lambda_n+\mu_n}$$ for natural $n$ can be rewritten as $$u_{n+1}=u_n-\frac{\pi_n}{\mu_n}, $$ which implies $$u_n=u_1-\sum_{j=1}^{n-1}\frac{\pi_j}{\mu_j}. \tag{1} $$ 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 (1) imply $$\pi_n h_n=u_n=\sum_{j=n}^\infty\frac{\pi_j}{\mu_j}\to0 $$ as $n\to\infty$ iff $w_1<\infty$.

Finally, the desired conclusion $$\lim_{n\to\infty}\prod_{j=1}^n\frac{\lambda_j}{\mu_j}(w_n-w_{n+1)}=0$$ can be rewritten as $\lim_{n\to\infty}\pi_n h_n=0$. Thus, the desired conclusion holds iff $w_1<\infty$.

For natural $j$, let $$h_j:=w_j-w_{j-1},\quad\pi_j:=\prod_{i=1}^{j-1}\frac{\lambda_i}{\mu_i}, $$ so that $\pi_1=1$, and $$u_j:=\pi_j h_j.$$ Then the equation
$$w_n=\frac{\lambda_nw_{n+1}+\mu_nw_{n-1}+1}{\lambda_n+\mu_n}$$ for natural $n$ can be rewritten as $$u_{n+1}=u_n-\frac{\pi_n}{\mu_n}, $$ which implies $$u_n=u_1-\sum_{j=1}^{n-1}\frac{\pi_j}{\mu_j}. \tag{1} $$ 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 (1) imply $$\pi_n h_n=u_n=\sum_{j=n}^\infty\frac{\pi_j}{\mu_j}. $$ So, $\pi_n h_n\underset{n\to\infty}\longrightarrow0$ iff $w_1<\infty$.

Finally, the desired conclusion $$\lim_{n\to\infty}\prod_{j=1}^n\frac{\lambda_j}{\mu_j}(w_n-w_{n+1)}=0$$ can be rewritten as $\lim_{n\to\infty}\pi_n h_n=0$.

Thus, the desired conclusion holds iff $w_1<\infty$.