This is true. Write $w(x,y)=x^{m_1}y^{n_1}\ldots x^{m_k} y^{n_k}$, with $m_j,n_j\in\mathbb Z$.

Suppose that your sentence fails in some infinite group. So $\forall y\: w(a,y)\not= 1$ in this group, for some $a$. Then in particular, taking $y=a^r$, we have that $a^{m+rn}\not=1$ for all $r\in\mathbb Z$, with $m=\sum m_j$, $n=\sum n_j$. This implies that $m+zn=0$ has no solution $z\in\mathbb Z$. But then $n$ has a prime factor $p$ that does not occur in $m$, and thus $m+zn\equiv 0\mod p$ can not be solved in $\mathbb Z_p$ either. In other words, $\forall x\exists y\: w(x,y)=1$ already fails in the finite cyclic group of order $p$.

This is true. Write $w(x,y)=x^{m_1}y^{n_1}\ldots x^{m_k} y^{n_k}$, with $m_j,n_j\in\mathbb Z$.

Suppose that your sentence fails in some infinite group. So $\forall y\: w(a,y)\not= 1$ in this group, for some $a$. Then in particular, taking $y=a^r$, we have that $a^{m+rn}\not=1$ for all $r\in\mathbb Z$, with $m=\sum m_j$, $n=\sum n_j$. This implies that $m+zn=0$ has no solution $z\in\mathbb Z$. But then $n$ does not divide $m$, and thus $m+zn\equiv 0\mod n$ can not be solved in $\mathbb Z_n$ either, or we have $n=0$, $m\not=0$. In either case, $\forall x\exists y\: w(x,y)=1$ already fails in a finite cyclic group (of order $n$, if $n\not=0$).