The condition $f(x + 1019) = f(x)$ allows to make all computations modulo $1019$, which is a prime number. Hence the problem asks the number of multiplicative functions in $\mathbb{Z}/1019\mathbb{Z}$.

Two such functions are $f(x) \equiv 1 \pmod{1019}$ and $f(x) \equiv 0 \pmod{1019}$. From now on suppose $f$ is different from these two functions.

Substituting $y = 0$, $f(0) = f(x)f(0) \Leftrightarrow f(0) = 0$; substituting $y = 1$, $f(x) = f(x)f(1) \Leftrightarrow f(1) = 1$. By Euler-Fermat's theorem, $x^{\phi(1019)} \equiv 1 \pmod{1019}$ so $f(x)^{\phi(1019)} = f(x^{\phi(1019)}) = f(1) = 1 \Leftrightarrow f(x) = \pm 1$.

Now let $g$ be a primitive root of $1019$ (that is, $g$ such that the least positive integer $d$ such that $g^d \equiv 1 \pmod{1019}$ is $\phi(1019)$). So every $x$ in $\mathbb{Z}/1019\mathbb{Z}$ can be written as $g^k$ for some $k$. Then $f(x) = (f(g))^k$. If $f(g) = 1$ then $f(x) = 0$ if $1019$ divides $x$ and $1$ otherwise; if $f(g) = -1$ then $f(x) = 0$ if $1019$ divides $x$, $f(x) = 1$ if $x$ is a quadratic residue modulo $1019$ (such residues are $1, g^2, g^4, \dots, g^{1016}$) and $f(x) = -1$ otherwise.