I wouldn’t use the pigeonhole principle: it’s not needed. You know that there is a unique integer $n$ such that $0\le n and $\frac{n}m\le\{x\}<\frac{n+1}m$. Now consider $x+\frac{k}m$, where $0\le k:
$x+\frac{k}m=\lfloor x\rfloor +\{x\}+\frac{k}m\;,$ so
$\lfloor x\rfloor+\frac{n+k}m\le x+\frac{k}m<\lfloor x\rfloor +\frac{n+1+k}m\;.\tag{1}$
What does $(1)$ tell you about $\left\lfloor x+\frac{k}m\right\rfloor$? Specifically, for how many values of $k$ will it be $\lfloor x\rfloor$, and for how many will it be $\lfloor x\rfloor+1$? If you can answer that, you’ll have a good handle on the righthand side of your desired equation.
As for the lefthand side, start from the fact that $\lfloor mx\rfloor=\big\lfloor m(\lfloor x\rfloor+\{x\})\big\rfloor$.
Most of the Missing Detail: If $0\le k\le m-n-1$, $(1)$ tells us that
$\lfloor x\rfloor\le\lfloor x\rfloor+\frac{n}m\le \lfloor x\rfloor+\frac{n+k}m\le x+\frac{k}m<\lfloor x\rfloor+\frac{n+k+1}m=\lfloor x\rfloor+1\;;$
after getting rid of the excess baggage, we have $\lfloor x\rfloor\le x+\frac{k}m<\lfloor x\rfloor+1\;,$ and therefore $\left\lfloor x+\frac{k}m\right\rfloor=\lfloor x\rfloor\;.\tag{2}$
On the other hand, if $m-n\le k\le m-1$, $(1)$ implies that
$\lfloor x\rfloor+1\le\lfloor x\rfloor+\frac{n+k}m\le x+\frac{k}m<\lfloor x\rfloor+\frac{n+1+k}m\le\lfloor x\rfloor+\frac{n+m}m<\lfloor x\rfloor+2$
and hence that $\left\lfloor x+\frac{k}m\right\rfloor=\lfloor x\rfloor +1\;.\tag{3}$
Now $(2)$ holds for $m-n$ values of $k$, and $(3)$ holds for the other $n$ values of $k$, so
$\begin{align*}\sum_{k=0}^{m-1}\left\lfloor x+\frac{k}m\right\rfloor&=(m-n)\lfloor x\rfloor+n(\lfloor x\rfloor+1)\\ &=m\lfloor x\rfloor+n\;. \end{align*}$
But $\begin{align*} \lfloor mx\rfloor&=\big\lfloor(m\lfloor x\rfloor+\{x\})\big\rfloor\\ &=\big\lfloor m\lfloor x\rfloor+m\{x\}\big\rfloor\\ &=m\lfloor x\rfloor+\lfloor m\{x\}\rfloor\;, \end{align*}$
since $m\lfloor x\rfloor$ is an integer.
Now you need only show that $\lfloor m\{x\}\rfloor=n$ to finish the argument. Go back to the beginning to recall how $n$ was defined.