Here is a slight modification of YCor's solution, which
is too long to describe in a comment. It is proved in the same way.

**Claim.** Any identity $w(x_1,\ldots,x_n)\approx 1$
which hold almost everywhere
in an infinite group must hold everywhere.

Here, an $n$-ary identity $w\approx 1$ holds almost everywhere
in infinite $G$ means the solution set $S\subseteq G^n$
of $w(x_1,\ldots,x_n) = 1$
satisfies $|G^n-S|<|G^n|=|G|$.

Step 1. If $w\approx 1$ holds almost everywhere, then so does
$w(x,x,x,x,\ldots,x) \approx 1$, and this has the form
$x^k=1$ for some $k$ (possibly $k=0$). As noted
YCor's comment to his solution, this
implies $x^k=1$ holds everywhere.
Thus we may assume that $w(x,x,\ldots,x)\approx 1$ holds everywhere.

Step 2. If $w\approx 1$ did not hold everywhere, then there would
exist a tuple $t=(g_1,\ldots,g_n)\in G^n$ that does not satisfy it.
Each conjugate of $t$ fails $w\approx 1$, so the
index of the centralizer of $t$ is small, forcing
$|C_G(t)|=|G|$.

Step 3. For each $h\in C_G(t)$ we have
$$
w(hg_1,hg_2,\ldots,hg_n) = w(h,h,\ldots,h)
w(g_1,g_2,\ldots,g_n) = 1\cdot w(g_1,g_2,\ldots,g_n) \neq 1,
$$
yielding $|G|$-many failures of $w\approx 1$,
namely all tuples in $C_G(t)\cdot t$. This is too many failures
of $w\approx 1$, thereby contradicting the existence of even one failure $t$ of $w\approx 1$.