Affine embedding of the two-point lattice into a semiring-like structure A commutative semiring-like structure is a structure $(R, {+}, {\cdot})$ where $+$ and $\cdot$ are associative and commutative, and $\cdot$ distributes over $+$.
Is there a commutative semiring-like $R$ such that we can embed the two element lattice $\{\bot, \top\}$ into $R$ so that meet and join are affine mappings? More precisely, we seek a semirig $R$ with distinct elements $F, T \in R$ such that, for some $a, b, c \in R$ we have the "truth table for disjunction"
\begin{align*}
a \cdot F + b \cdot F + c &= F \\
a \cdot F + b \cdot T + c &= T \\
a \cdot T + b \cdot F + c &= T \\
a \cdot T + b \cdot T + c &= T
\end{align*}
and there are some $d, e, f \in R$ that give us the "truth table for conjunction":
\begin{align*}
d \cdot F + e \cdot F + f &= F \\
d \cdot F + e \cdot T + f &= F \\
d \cdot T + e \cdot F + f &= F \\
d \cdot T + e \cdot T + f &= T ?
\end{align*}
A cursory search reveals that no such structure of size 2, 3, 4 or 5 exists.
Supplemental: Ludwig Maes, who originally asked me this question, observes that
\begin{align*}
  F + F &= (d \cdot F + e \cdot T + f) + (d \cdot T + e \cdot F + f) \\
        &= (d \cdot T + e \cdot T + f) + (d \cdot F + e \cdot F + f) \\
        &= T + F.
\end{align*}
Thus, if there is such a structure, it can't have good cancellation properties, or at least $F$ and $T$ should not, or else we get $F = T$. This rules out all rings, for example.
Additional question: It turns out the original question did not assume commutativity of $\cdot$ (in which case we assume both the left and right distributivity of $\cdot$ over $+$). It would be interesting to have an answer also for the non-commutative case. Unfortunately, Keith Kearnes's beautiful use of the majorit function does not work without commutativity of $\cdot$.
 A: There is no commutative semiring with the desired
properties.
To see this, suppose instead that the semiring
$R$ contains a subset $\{F, T\}$ supporting affine operations
$\vee, \wedge$ that are lattice operations on $\{F, T\}$.
$R$ then has 
an affine operation
$m(x,y,z) := (x\wedge y)\vee (x\wedge z)\vee (y\wedge z)$ that is a majority operation on $\{F, T\}$.
Let $m(x,y,z)=A_1x+A_2y+A_3z+B$ be an affine representation for
this operation.
By comparing coefficients and constant terms, observe that
$$
m(m(x_{11},x_{12},x_{13}),m(x_{21},x_{22},x_{23}),m(x_{31},x_{32},x_{33}))\\
=
m(m(x_{11},x_{21},x_{31}),m(x_{12},x_{22},x_{32}),m(x_{13},x_{23},x_{33}))
$$
holds throughout $R$.
(Specifically, the coefficient of $x_{ij}$ on either side is $A_iA_j$,
while the constant term on either side is $A_1B+A_2B+A_3B+B$.)
But under the substitution
$x_{11}=x_{12}=x_{22}=x_{23}=T$ and
$x_{13}=x_{21}=x_{31}=x_{32}=x_{33}=F$
the left side gives $T$ while the right side gives $F$ (since $m(x,y,z)$
acts like majority on $\{F, T\}$).
This is a contradiction. $\Box$

edit 6/23/17.
The question was edited to ask
Is there a noncommutative semiring $R$
such that we can embed the 2-element lattice
into $R$ so that meet and join are affine mappings?
Let me explain how to convert
this question to a more tractable one, and then I will
answer the tractable version.

Conversion:
Part 1.
An algebra $A$ is embeddable in a semiring
so that its operations have affine
polynomial representations iff
$A$ is embeddable in a semimodule
so that its operations have 
polynomial representations. (Here to embed $A$
means to represent $A$
as a subalgebra of a reduct of the polynomial
expansion.)
[Reasoning: If $A$ is affinely embeddable in a semiring $R$,
view $R$ as a semimodule over itself to get $A$ embedded
in a semimodule. For the other direction, if $A$
is embedded in an $S$-semimodule $M$, can affinely embed
$A$ in the matrix semiring 
$\left[\begin{matrix} S&M\\0&0\end{matrix}\right]$.]
Part 2.
The 2-element lattice is embeddable in a semimodule
iff the 2-element majority algebra is embeddable.
[Reasoning:
If $\{T,F\}$ supports lattice polynomials $\vee, \wedge$,
then it supports a majority polynomial
$m(x,y,z) = (x\vee y)\wedge (x\vee z)\wedge (y\vee z)$.
Conversely, if $\{T,F\}$ supports a majority polynomial
$m(x,y,z)$, then it supports lattice polynomials
$x\vee y = m(x,y,T)$ and $x\wedge y = m(x,y,F)$.]

I reformulate the edited question as:
Is there a semimodule $M$ over a noncommutative
semiring $R$
such that we can embed the 2-element majority
algebra into $M$ as a subalgebra of a reduct
of the polynomial expansion?
Answers: 
(1) Yes, it is possible
to embed the 2-element majority algebra into a semimodule,
but (2) not into a finite semimodule.
The explanation of why it is possible
to embed the 2-element majority algebra into a semimodule
can be found in 
Jaroslav Jezek,
Terms and semiterms,
Commentationes Mathematicae Universitatis Carolinae
20 (1979), 447-460.
In this paper it is shown that every algebra is embeddable
in a semimodule.
(Hence every algebra is affinely embeddable in a semiring.)

However, it is not possible to embed the 2-element majority
algebra into a finite semimodule. (Equivalently,
it is not possible to embed the
2-element lattice into a finite
semiring so that its operations have affine
polynomial representations.)
This can be proved by modifying my proof for the commutative case.
The modification is a bit long, but I will include it
here for those who want to check.
Proof (That you can't embed the 2-element majority
algebra into a finite semimodule.)
Assume that $M$ is a finite semimodule over the semiring
$R$, which we may assume acts faithfully on $M$.
The faithfulness assumption forces $R$ to be finite as well.
Assume also that the semimodule polynomial
$m(x,y,z) = \alpha x + \beta y + \gamma z + d$ acts like
a majority operation on
the set $\{T, F\}\subseteq M$.
Call $m^{(2)}(x,y,z):=
m(m(x,y,z),y,z)$ the 2nd first-variable iterate of $m$,
$m^{(3)}(x,y,z):=m(m(m(x,y,z),y,z),y,z)$ the 3rd
first-variable iterate of $m$, ETC. Each $m^{(k)}(x,y,z)$
is a majority polynomial on $\{T,F\}$, and has the form
$\alpha^{(k)} x + \beta^{(k)} y + \gamma^{(k)} z + d^{(k)}$.
I may select any one of these to be my
majority polynomial as I continue the argument.
Since we are iterating in the first variable, $\alpha^{(k)} = \alpha^k$.
Since $M$ is a semimodule over a finite semiring,
it is possible to choose $k$ so that $\alpha^{2k}=\alpha^k$.
This allows me to replace the original majority polynomial with 
some first variable iterate, change notation back,
and henceforth assume that the majority polynomial
$m(x,y,z) = \alpha x + \beta y + \gamma z + d$
was selected to satisfy $\alpha^2=\alpha$.
(This is where the finiteness is used!)
Now $T=m(T,F,T)=\alpha T + (\beta F + \gamma T + d)$ is not equal to
$F=m(F,F,T)=\alpha F + (\beta F + \gamma T + d)$, so 
necessarily $\alpha T\neq \alpha F$.
Let $T' = \alpha T$ and $F' = \alpha F$. I claim that

* there are inverse semimodule polynomial bijections
$f\colon\{T, F\}\to \{T',F'\}$ and
$g\colon\{T', F'\}\to \{T,F\}$,

* these bijections allow us to conjugate $m$ to a majority
polynomial $\mu (x,y,z):=f(m(g(x),g(y),g(z))$ on $\{T', F'\}$, 

* $\mu (x,y,z)$ can be further modified to a majority 
polynomial $\overline{\mu}(x,y,z)$
on $\{T',F'\}$ where the coefficient of $x$ commutes with the
coefficients of $y$ and $z$, and finally

* this is enough commutativity to make the proof
for the commutative case work here.
To establish the first bulleted item,
let $f(x)=\alpha x$ and $g(x)=x+\beta F+\gamma T+d$.
That $f\colon \{T, F\}\to \{T',F'\}$ is a bijection follows from the
definitions of $T'$ and $F'$. That $g\colon\{T', F'\}\to \{T,F\}$
is the inverse follows from the majority equations for $m$.
The second bulleted item follows from the fact that conjugation
preserves the majority identities.
For the third bulleted item, it is easy to see that the coefficient
of $x$ in $\mu (x,y,z) = f(m(g(x),g(y),g(z))$ is $\alpha^2 = \alpha$.
That is, $\mu (x,y,z) = \alpha x + (\textrm{some stuff})$.
If it helps to write it out, we have
$$
\mu(x,y,z) = \alpha x + \alpha \beta y + \alpha\gamma z + D
$$
where
$$
\begin{array}{rl}
D &= \alpha\beta F + \alpha\gamma T + \alpha d + \alpha\beta^2 F
+\alpha\beta\gamma T + \alpha\beta d\\
&+ \alpha\gamma\beta F + \alpha\gamma^2 F
\alpha\gamma^2 T+\alpha\gamma d + \alpha d.
\end{array}
$$
Let $\overline{\mu}(x,y,z)=x + \alpha\beta y + \alpha\gamma z + D$.
That is, delete the coefficient
$\alpha$ from $x$ in the polynomial expression
for $\mu(x,y,z)$ (or think of it
as replacing $\alpha$ with $1$).
Observe that the semimodule
polynomials $\mu (x,y,z)$ and $\overline{\mu}(x,y,z)$
both have the same restriction to $\{T',F'\} = \{\alpha T, \alpha F\}$,
since the polynomials only differ in their $x$-coefficient,
the inputs all have $\alpha$ as a prefix, and $\alpha^2 = \alpha$.
For example, 
$$
\begin{array}{rl}
\overline{\mu}(F',T',T')&=
\overline{\mu}(\alpha F,\alpha T,\alpha T)\\
&=(\alpha F) +
\alpha\beta (\alpha T) + \alpha\gamma (\alpha T) + D\\
&= 
\alpha(\alpha  F) +
\alpha\beta (\alpha T) + \alpha\gamma (\alpha T) + D\\
&= 
\mu(\alpha F,\alpha T,\alpha T)\\
&=
\mu(F',T',T').
\end{array}$$
[Stock-taking:]
We started with the assumption that a $2$-element subset
$\{T, F\}$ of $M$ supports a majority polynomial $m(x,y,z)$,
and constructed a new instance $\{T', F'\}$
and $\overline{\mu}(x,y,z)$, but in the latter instance
the coefficient of $x$ commutes with the coefficients of $y$ and $z$
(since in the latter case the coefficient of $x$ is $1$).
Now, modifying the proof in the commutative case, we have
$$
\overline{\mu}
(\overline{\mu}(x_{11},x_{12},x_{13}),\overline{\mu}(x_{21},x_{22},x_{23}),\overline{\mu}(x_{31},x_{32},x_{33}))\\
=
\overline{\mu}
(\overline{\mu}(x_{11},x_{21},x_{31}),\overline{\mu}(x_{12},x_{22},x_{23}),\overline{\mu}(x_{13},x_{32},x_{33})).
$$
But under the substitution $x_{11}=x_{12}=x_{13}=x_{23}=x_{33}=T'$
and $x_{21}=x_{22}=x_{31}=x_{32}=F'$
the left side gives $F'$ while the right side gives $T'$.
$\Box$
