= Solution
In this <localization of a ring>, a nonzero fraction $a/b$ with odd $b$ is a <unit> exactly when $a$ is odd: its reciprocal then has odd denominator; if $a$ is even, an inverse would equate an even integer with an odd one after clearing denominators. Consequently every nonzero element has a unique form
$$
x=2^r u,\qquad r\geq0,\quad u\in R^\times.
$$
Here $r$ is the exponent of $2$ in the numerator of the reduced fraction, its nonnegative <2-adic valuation>.
Every nonzero <ideal> has the form $dR$ by part (c). Write the nonzero integer $d=2^r v$ with $v$ odd; since $v$ is a <unit>, $dR=2^rR$. Hence the nonzero proper <ideals> are exactly
$$
\boxed{2^rR=\{a/b\in R:2^r\mid a\},\qquad r=1,2,3,\ldots.}
$$
The <zero ideal> and $R=2^0R$ complete the full list. These ideals are distinct: $2^r$ does not belong to $2^{r+1}R$, since $1/2\notin R$.
The reduction map $a/b\mapsto a\bmod2$ is a well-defined surjective <ring homomorphism> $R\to\mathbb F_2$ with <kernel of a ring homomorphism> $2R$; odd denominators reduce to $1$. Thus $R/2R\cong\mathbb F_2$ is a <field>, so $2R$ is a <prime ideal> and $2$ is a <prime element>. Every nonzero <nonunit> is associated to $2^r$ with $r\geq1$. If $r\geq2$, the factorization $2^r=2\cdot2^{r-1}$ is into two <nonunits>, so it is not an <irreducible element> and therefore not a <prime element>. We conclude
$$
\boxed{\text{The sole prime element up to units is }2.}
$$
The failed step in the argument of part (b) is the guarantee that the product plus one is a nonunit and hence has an irreducible factor. Here every <prime element> is an even fraction, so a product of <prime elements> plus $1$ is an odd fraction, therefore a <unit>. Already $2+1=3$ has no prime divisor in this ring. In $k[t]$, positive degree prevented that failure.
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