Assume that $x_n$ converges to a nonzero number $x$ and that the sum $x_ny_n$ converges to a limit $L$. Prove that the series $y_n$ converges.
The natural guess is that $y_n$ will converge to $L/x$. What I did was $$ \begin{align} |x|\;\left|y_n-\frac{L}{x}\right|&=|xy_n-L| \\ &=|xy_n-L+x_ny_n-x_ny_n|\\ &=|(x_ny_n-L)+(xy_n-x_ny_n)|\\ &\leq |x_ny_n-L|+|x_ny_n-xy_n| \\ &=|x_ny_n-L|+|y_n|\;|x_n-x| \end{align} $$ Then using the converge of $x_n$ and $x_ny_n$, one can find the appropriate $N$'s given any $\epsilon>0$ for the terms on the right$-$so long as $y_n$ is bounded. But as $x_ny_n$ converges, it is bounded. Then there is an $M$ such that $$ M\geq |x_ny_n|=|x_n|\;|y_n| $$ for all $n$. But this of course implies that $$ \frac{M}{|x_n|} \geq |y_n| $$ (where $x_n \neq 0$, of course it cannot be zero everywhere as $x_n$ has a nonzero limit and we can 'ignore' any other possible zero issues) so that $y_n$ is bounded.
Does this look sufficient? I feel there should be a much simpler shorter way to do this (without using anything 'fancy').
You mean sequences not series. A seemingly natural analog of this result for series is in fact false: multiplying a convergent series elementwise by a convergent sequence can result in a divergent series.
The result is really just a restatement of the Limit of a Quotient Theorem: if $a_n\to a$ and $b_n\to b\ne 0$, then $a_n/b_n \to a/b$. If you have this theorem at your disposal, then just use it with $a_n=x_ny_n$ and $b_n=x_n$. If not, then you better prove the Limit of a Quotient Theorem first, which has a more natural workflow and is a standard result anyway.