This is a consequence of one of the basic limit laws:
If $\lim\limits_{t\to a} F(t)$ exists and $\lim\limits_{t\to a}G(t)$ exists, then $\lim\limits_{t\to a}\Bigl( F(t)-G(t)\Bigr)$ exists, and $\lim_{t\to a}\Bigl(F(t)-G(t)\Bigr) = \lim_{t\to a}F(t) - \lim_{t\to a}G(t).$
So, we are assuming that $\lim_{h\to 0}\frac{f(1+h)}{h}\ \text{exists.}$ We are also assuming that $f(x)$ has a derivative at $1$; that means that $\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}\ \text{exists.}$
Therefore, $\lim_{h\to 0}\left( \frac{f(1+h)}{h} - \frac{f(1+h)-f(1)}{h}\right)\ \text{exists}$ and moreover, $\lim_{h\to 0}\left(\frac{f(1+h)}{h} - \frac{f(1+h)-f(1)}{h}\right) = \lim_{h\to 0}\frac{f(1+h)}{h} - \lim_{h\to 0}\frac{f(1+h)-f(1)}{h}$ by the limit law quoted above, with $F(h) = \frac{f(1+h)}{h}$ and $G(h) = \frac{f(1+h)-f(1)}{h}$.
Now just notice that $\begin{align*} \frac{f(1+h)}{h} - \frac{f(1+h)-f(1)}{h} &= \frac{f(1+h)-(f(1+h)-f(1))}{h}\\ &= \frac{f(1+h)-f(1+h)+f(1)}{h}\\ &= \frac{f(1)}{h}, \end{align*}$ giving the equality you have when you put everything together.
That means that \begin{align*} \lim_{h\to 0}\frac{f(1)}{h} &= \lim_{h\to 0}\left(\frac{f(1+h)}{h} - \frac{f(1+h)-f(1)}{h}\right)\\ &= \left(\lim_{h\to 0}\frac{f(1+h)}{h}\right) - \left(\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}\right)\\ &= 1 - f'(1), \end{align*} since your assumption is that the first limit equals $1$, and by definition the second limit is the derivative at $1$.