If $f(a+b)=f(a)f(b)$, $f(0)=1$, and $f$ is differentiable at $x=0$, show that $f$ is differentiable for all $x$, and that $f'(x)=f'(0)f(x)$

If what is bothering you is that you seem to be assuming that $f$ is differentiable at $x$ in your proof, you can rewrite the proof as follows. Note that for all $x,h\in\mathbb R$ such that $h\neq0$, $$ \frac{f(x+h)-f(x)}{h}=\frac{f(x)f(h)-f(x)}{h}=f(x)\cdot \frac{f(h)-1}{h}=f(x)\cdot\frac{f(h)-f(0)}{h} \, . $$ Now, we can use the "product law" for limits: if $\lim_{x\to a}f(x)$ exists and equals $l$, and $\lim_{x\to a}g(x)$ exists and equals $m$, then $\lim_{x\to a}f(x)g(x)=lm$. By assumption, $f$ is differentiable at $0$, meaning that $$ f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h} $$ exists. Moreover, $f(x)$ is a constant with respect to $h$, and so $\lim_{h\to0}f(x)=f(x)$. Therefore, using the "product law", we find that $$ \lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}f(x)\cdot\frac{f(h)-f(0)}{h} $$ exists, and equals $f(x)f'(0)$. That's it.

Since $f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$ exists we have $f'(0)=f(0)\cdot\lim_{h\to 0}\frac{f(h)-1}{h}$ and thus $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f(x)\cdot\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\frac{f'(0)}{f(0)}\cdot f(x)$$ and therefore $f$ is differentiable everywhere with $f'(x)=f'(0)\cdot f(x)$.

There are a couple of other interesting things about this function: $$f(a+b)=f(a)f(b)\qquad f(0)=1$$ $$1=f(0)=f(1-1)=f(-1)f(1)$$ and you can apply this in general: $$1=f(0)=f(x-x)=f(x)f(-x)\Rightarrow f(-x)=\cfrac{1}{f(x)}$$