For a positive integer $N$, write $N=2^k\left(\prod_ip_i^{l_i}\right)\left( \prod_jq_j^{m_j}\right)$, where the $p_i$ and $q_j$ are primes with $p_i\equiv1\pmod{4}$ and $q_j\equiv3\pmod{4}$. The proportion of the grid that an $N$-mover can reach is
$$f(N)=\left\{\begin{array}{ll}
0&\text{ if } m_j\ \text{ odd for some } j\\
2^{-k}\prod_jq_j^{-m_j}&\text{ otherwise}
\end{array}\right.,$$
so your conjecture is correct.
The proof below is by induction on the prime factors of $N$

In the base case $N=1$ clearly $f(N)=1$ as the $N$-mover can make the moves $(1,0)$ and $(0,1)$.

If the $N$-mover can make the move $(u,v)$ then clearly the $4N$-mover can make the move $(2u,2v)$. Conversely, if the $4N$-mover can make the move $(r,s)$ then
$$4N=r^2+s^2,$$
and reducing mod $4$ shows that both $r$ and $s$ are even, say $r=2u$ and $s=2v$. Then
$$u^2+v^2=\frac{r^2+s^2}{4}=N,$$
so the $4N$-mover can make the move $(r,s)$ if and only if the $N$-mover can make the move $(u,v)$, which shows that $f(4N)=\frac{1}{4}f(N)$. To prove that $f(2^kN)=2^{-k}f(N)$ for all $N$ and $k\geq0$ it now suffices to verify that $f(2N)=\frac{1}{2}f(N)$ for odd $N$.

If $N$ is odd and the $N$-mover can make the move $(u,v)$ then
$$(u+v)^2+(u-v)^2=2u^2+2v^2=2N,$$
so the $2N$-mover can make the move $(u+v,u-v)$. Conversely, if the $2N$-mover can make the move $(r,s)$, then
$$2N=r^2+s^2,$$
which implies that both $r$ and $s$ are odd, so in particular $\frac{r+s}{2}$ and $\frac{r-s}{2}$ are integers. Note that
$$\left(\frac{r+s}{2}\right)^2+\left(\frac{r-s}{2}\right)^2=\frac{r^2+s^2}{2}=N,$$
so the $N$-mover can make the move $\left(\tfrac{r+s}{2},\tfrac{r-s}{2}\right)$. This yields a bijection between the points the $N$-mover can reach and the points the $2N$-mover can reach. It is in fact a linear transformation with determinant $-2$ and so $f(2N)=\frac{1}{2}f(N)$, as desired.

Primes $q\equiv3\pmod{4}$ allow a similar argument. If the $qN$-mover can make the move $(r,s)$ then
$$qN=r^2+s^2,$$
and reducing mod $q$ shows that $r^2+s^2\equiv0\pmod{q}$. Because $q\equiv3\pmod{4}$ it follows that $r\equiv s\equiv0\pmod{q}$, say $r=qu$ and $s=qv$, and hence we have
$$u^2+v^2=\frac{r^2+s^2}{q^2}=\frac{N}{q},$$
so in particular $q\mid N$. This shows that $f(qN)=0$ if $q\nmid N$. Of course, if the $\frac{N}{q}$-mover can make the move $(u,v)$ then the $qN$-mover can make the move $(qu,qv)=(r,s)$, which proves that $f(q^2N)=q^{-2}f(N)$, and hence also that $f(N)=0$ if the highest power of $q$ dividing $N$ is odd.

It remains to show that $f(N)=1$ for integers $N$ that are a product of primes congruent to $1\pmod{4}$. For this the following lemma is convenient:

**Lemma:** *Let $a,b\in\Bbb{Z}$ with $a$ even and $b$ odd and let $d:=\gcd(a,b)$. If the $N$-mover can make the move $(a,b)$, then it can reach $(d,0)$.*

*Proof.* Let $a':=\frac{a}{2}$ and $b':=\frac{b-1}{2}$. Then the $N$-mover can reach
\begin{eqnarray*}
a'(a,b)+a'(-a,b)&=&a'(0,2b)=(0,ab)\\
b'(b,a)+b'(-b,a)&=&b'(0,2a)=(0,a(b-1)),
\end{eqnarray*}
and hence also $(0,ab)-(0,a(b-1))=(0,a)$. By symmetry it can also reach $(a,0)$ and so it can reach $(a,0)+(-a,b)=(0,b)$ from which the conclusion follows.$\hspace{10pt}\square$

Let $N$ be a product of primes congruent to $1$ mod $4$ and let $p$ be a prime with $p\equiv1\pmod{4}$. Then $p=x^2+y^2$ for coprime integers $x$ and $y$. If the $N$-mover can make the move $(u,v)$ then $u^2+v^2=N\equiv1\pmod{4}$. Without loss of generality $x$ and $u$ are even and $y$ and $v$ are odd. Then
$$pN=(x^2+y^2)(u^2+v^2)=(xu\pm yv)^2+(yu\mp xv)^2,$$
for both choices of opposite signs. Hence the $pN$-mover can make the moves $(xu\pm yv,yu\mp xv)$, where $xu\pm yv$ is odd and $yu\mp xv$ is even. Then by the lemma, the $pN$-mover can reach $(z_{\pm},0)$ where $z_{\pm}:=\gcd(xu\pm yv,yu\mp xv)$, and hence it can reach $(z,0)$ where
$$z:=\gcd(z_+,z_-)=\gcd(xu+yv,yu-xv,xu-yv,yu+xv).$$
[Thanks to Ørjan Johansens comments:]
Clearly $z$ is odd, and $z\mid2\gcd(u,v)$ because
$$(xu+yv)+(xu-yv)=2xu
\qquad\text{ and }\qquad
(yu+xv)+(yu-xv)=2yu,$$
$$(xu+yv)-(xu-yv)=2yv
\qquad\text{ and }\qquad
(yu+xv)-(yu-xv)=2xv,$$
where $\gcd(x,y)=1$. It follows that $z\mid\gcd(u,v)$ and hence that the $pN$-mover can reach $(u,v)$. This shows that $f(pN)\geq f(N)$, and so by induction that $f(N)=1$ for every integer $N$ that is a product of primes congruent to $1$ mod $4$, completing the proof.