It is possible to construct a translation-invariant strict extension of Lebesgue measure on $\mathbb{R}$.
Such a construction is sketched in Fremlin, Measure theory, Vol 4.I Exercise 442Yc, page 289:
Show that there is a set $A\subseteq[0,1]$, of Lebesgue
outer measure $1$, such that no countable set of translates of $A$
covers any set of Lebesgue measure greater than $0$.
Hint: Let
$\langle F_{\xi}\rangle_{\xi<\frak c}$ run over the uncountable closed
subsets of $[0,1]$ with cofinal repetitions (4A3Fa), and enumerate the
countable subsets of $\Bbb R$ as $\langle I_{\xi}\rangle_{\xi<\frak c}$.
Choose inductively $x_{\xi}$, $x'_{\xi}\in F_{\xi}$ such that
$x_{\xi}\notin\bigcup_{\eta,\zeta<\xi}x'_{\eta}-I_{\zeta}$,
$x'_{\xi}\notin\bigcup_{\eta,\zeta\le\xi}x_{\eta}+I_{\zeta}$; set
$A=\{x_{\xi}:\xi<\frak c\}$.
Show that we can extend Lebesgue measure
on $\Bbb R$ to a translation-invariant measure for which $A$ is
negligible.
Hint: 417A.
The Lemma in 417A reads
Let $(X,\Sigma,\mu)$ be a semi-finite measure
space, and ${\cal A}\subseteq{\cal P}X$ a family of sets such that
$\mu_*(\bigcup_{n\in\Bbb N}A_n)=0$ for every sequence $\langle A_n \rangle_{n\in\mathbb{N}}$
in $\cal A$. Then there is a measure $\mu'$ on $X$, extending $\mu$,
such that
(i) $\mu'A$ is defined and zero for every $A\in\cal A$,
(ii) $\mu'$ is complete if $\mu$ is,
(iii) for
every $F$ in the domain $\Sigma'$ of $\mu'$ there is an $E\in\Sigma$
such that $\mu'(F \mathbin{\Delta} E)=0$.}
In particular, $\mu$ and $\mu'$ have
isomorphic measure algebras, so that $\mu'$ is localizable if $\mu$ is.