Sum of reciprocals of rough numbers

(I am going to switch your $x$ and $y$ for aesthetic purposes.)

The correct upper bound is $\log x/\log y$, which in particular is bounded if $y$ is a power of $x$.

Let $\Phi(x,y)$ be the number of $y$-rough integers up to $x$. By integration by parts, your sum is $$\sum_{\substack{n \le x \\ n \text{ is }y\text{-rough}}}\frac{1}{n} = \int_{1^-}^{x} \frac{d\Phi(t,y)}{t} = \frac{\Phi(x,y)}{x}+\int_{1}^{x} \frac{\Phi(t,y)}{t^2}dt .$$

We have the classical bound $\Phi(t,y) \ll \frac{t}{\log y}$ if $y \le t$, and otherwise $\Phi(t,y)=1$. It follows that, for $y \le x$, $$\sum_{\substack{n \le x \\ n \text{ is }y\text{-rough}}}\frac{1}{n} \ll \frac{1}{\log y}+\frac{1}{\log y}\int_{y}^{x} \frac{dt}{t} + \int_{1}^{y} \frac{dt}{t^2} \ll \frac{\log(x/y)+1}{\log y} \ll u $$ where $u:=\log x/\log y$ as usual.

A lower bound of $\gg u$ can be exhibited as well. Indeed, just use $\Phi(t,y)\gg \frac{t}{\log y}$ for $y \le \sqrt{t}$ in the argument above to obtain that your sum is $\gg \frac{\log(x/y^2)}{\log y} =u-2 \gg u$ if $u \ge 3$. If $u<3$ one has the trivial lower bound $1$ (since $1$ is $y$-rough) which is $\gg u$.

Some comments:

• For $x\ge y >\sqrt{x}$, your sum is just a sum over primes and the number $1$ (which is $y$-rough for every $y$), as you observed, in which case Mertens' Theorem implies that your sum is $1+\sum_{y\le p \le x} \frac{1}{p} = 1+\log \log x - \log \log y + o(1) = 1+ \log u +o(1)$, which is bounded. For $y>x$, your sum is identically $1$.
• For $y=x^{o(1)}$ I believe the asymptotics can be shown to be $\prod_{p<y}(1-\tfrac{1}{p}) \log x$ (in the above proof use $\Phi(t,y)\sim t\prod_{p<y}(1-\tfrac{1}{p})$ for $y=t^{o(1)}$, which is a consequence of the fundamental lemma of sieve theory, say). If this $y$ also satisfies $y\to \infty$ then this is $\sim e^{-\gamma} u$ by Mertens' Theorem.
• For bounded $u=\log x/\log y \ge 2$, one should be able to show that the asymptotics is of the shape $u f(u)$ where $f$ satisfies a delay differential equation and $\lim_{u\to \infty} f(u)=e^{-\gamma}$. This can be proved by using the precise asymptotics for $\Phi(x,y)$ in terms of the Buchstab function.