Solution:
Let us first prove the result for nonnegative $y$ by strong induction on $y$. If $y=0$, (2) becomes
$$f(x)+f(x)=f(x) f(0),$$
which is true since $f(0)=2$, and if $y=1$, (2) is the same as (1). Let us assume (2) for $y=z-1$ and $y=z$ and try to prove it for $y=z+1$. We have
$$\begin{gathered} f(x+z-1)+f(x-z+1)=f(x) f(z-1) \\ f(x+z)+f(x-z)=f(x) f(z) \end{gathered}$$
Multiplying (4) by $f(1)$ and subtracting (3) yields
$$\begin{gathered} f(x+z) f(1)-f(x+z-1)+f(x-z) f(1)-f(x-z+1)=f(x)[f(z) f(1)-f(z-1)] \\ f(x+z+1)+f(x-z-1)=f(x) f(z+1) \end{gathered}$$
as desired.
It remains to prove (2) for $y<0$. If $y=-z$ and $z>0$, (2) becomes
$$f(x-z)+f(x+z)=f(x) f(-z)$$
which would follow from (2) for $y=z$ if $f(-z)=f(z)$. To prove this, let $x=0$ and $y=z$ in (2), getting
$$\begin{aligned} f(z)+f(-z) & =f(0) f(z) \\ f(z)+f(-z) & =2 f(z) \\ f(-z) & =f(z) . \end{aligned}$$