Suppose $X \sim Exponential(\lambda)$. That is, the PDF for $X$ is
$f_X(x)=\lambda \cdot e^{-\lambda x}$, $x\ge 0$, and the CDF of $X$ is
$F_X (x)=\int_{-\infty}^x f_X(x)=1-e^{-\lambda x}$, $x\ge 0$.
Then what can we say about the distribution of $\frac Xc$ ($c>0$)?
I think it is just $\frac Xc \sim Exponential(c\lambda)$, since we are looking for $F_{\frac Xc} (x)=Pr(\frac Xc \le x)= Pr(X\le cx)=F_X (cx)$. Is this correct reasoning? If so, wouldn't this work for any distribution, not just the exponential distribution?
You're nearly there, but not quite. You have to take the additional step to say that, if $Z=X/c$, then $$\tag{1}F_Z(x) = F_X(cx)=1-e^{-\lambda(cx)}=1-e^{-(c\lambda)x}=F_Y(x)$$ where $Y\sim\operatorname{Exp}(c\lambda)$, and therefore $Z$ and $Y$ are identically distributed.
More tersely, $$\tag{2}F_{\frac1c\operatorname{Exp}(\lambda)}(x)=F_{\operatorname{Exp}(\lambda)}(cx)=F_{\operatorname{Exp}(c\lambda)}(x)$$ so $$\tag{3}\frac1c\operatorname{Exp}(\lambda)\sim\operatorname{Exp}(c\lambda)$$
Note that it is the third equality in $(1)$ that is crucial here. Something like this isn't true in general, so you can't make the conclusion with just any distribution.