= Solution
Use the one-sided <Laplace transform> $F(s)=\int_0^\infty e^{-st}f(t)\,dt$. Since $f=0$ on $[0,1)$, substitution gives $\mathcal L[f(t+1)](s)=e^sF(s)$. Thus
$$
F(s)=\frac1{s(e^s-1/2)}=\frac{2e^{-s}}{s(1-e^{-s}/2)}.
$$
Its <poles> are $s=0$ and $s_n=-\log2+2\pi in$, with <residues> of $e^{st}F(s)$ respectively two and $2e^{s_nt}/s_n$. For $t>1$ close the <Bromwich inversion formula> to the left, using horizontal heights avoiding those <poles>. The left-side integrand decays like $e^{st}/s$, and on the right-side horizontal ends like $e^{s(t-1)}/s$, so the added integrals vanish with the usual symmetric limiting contour. The <residue theorem> gives
$$
\boxed{f(t)=2+2^{1-t}\sum_{n=-\infty}^{\infty}\frac{e^{2\pi int}}{2\pi in-\log2}\quad(t\geq1),}
$$
where the series is summed symmetrically and at jumps has the midpoint value.
For $t<1$ the vanishing can be checked directly in the inversion integral, rather than inferred from the recurrence. On any vertical line $\Re s=c>0$ expand the denominator geometrically, uniformly there:
$$
F(s)=2\sum_{j=0}^\infty2^{-j}\frac{e^{-(j+1)s}}s.
$$
Each inversion term is $2^{1-j}(2\pi i)^{-1}\int_{c-i\infty}^{c+i\infty}e^{s(t-j-1)}ds/s$. If $t<1$, close this contour to the right: the exponential decays, and the <pole> at zero is outside, so every term is zero. Equivalently the geometric expansion bounds the full closing contour by the same decaying factors. Thus \b[$f(t)=0$ for $t<1$].
The same integral gives $f(t)=2\sum_{j\geq0}2^{-j}H(t-j-1)$ elsewhere. For $m<t<m+1$ with <integer> $m\geq1$, this is $4(1-2^{-m})$, which directly verifies the difference equation. Neither a <Laplace transform> nor a Fourier series determines values at isolated jumps; the printed series fixes the standard convention $H(0)=1/2$, including $f(1)=1$.
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