A field is "two (abelian) groups in one".
Explicitly, $F(,+)$ is an abelian group, and $(F - \{0\},\cdot)$ is an abelian group.
We further require that the map $L_a: F \to F$ ("left multiplication by $a$") given by $L_a(b) = a\cdot b$ is an $(F,+)$-homomorphism. This is the familiar distributive law:
$L_a(b+c) = L_a(b) + L_a(c)$, or: $a\cdot (b+c) = a\cdot b + a\cdot c$.
This is a "compatibility" requirement, ensuring our two operations play nice together.
A ring is similar, we still require an additive abelian group. But now, we do not require that $(R - \{0\},\cdot)$ be an abelian group, rather only that $(R,\cdot)$ be a simpler structure, known as a semigroup.
We still require $L_a$ be additive (an additive group homomorphism). However, since semigroups are not, typically, commutative (abelian), we also have to make this requirement of $R_a: R \to R$, where $R_a(b) = b\cdot a$.
Life is easier for us, of course, if $(R,\cdot)$ is commutative, and easier still if $(R,\cdot)$ forms a "commutative semigroup with identity" (the proper term is "commutative monoid").
As in groups, the identity element of a monoid is unique. If the multiplicative semigroup of a ring $R$ forms a commutative monoid, we call $R$ a commutative ring with unity.
Any monoid $M$(whether in a ring, or not) has a group associated with it-its "group of units" (uusally written $U(M)$). These are the "invertible" elements of $M$, that is, those elements $u$ of $M$ for which there exists $v \in M$ such that:
$uv = vu = 1_M$.
It is easy to see that the group of units actually form a sub-monoid of $M$ (which is why we can make a group from them-we have closure: if $u,v \in U(M)$, then $uv$ has the inverse $v^{1}u^{-1}$ and so is likewise in $U(M)$).
In rings, all the "action" is in the multiplicative semigroup. Since rings without unity (where we have *just* a semi-group) can behave somewhat perversely, we usually insist rings have a multiplicative monoid (but this is a point of some contention, even amognst mathematicians).
In rings, one of our "goals" is to "divide as much as we can". We can't *always* divide, like we can in fields, but at least the ability to establish $a|b$ (that is: $b = ac$) let's us "break down" things into hopefully "simpler things" (N.B., this doesn't always work very well). In polynomial rings, this process is called "factoring".
Now the multiplicative identity of a field, $1_F$ is *still" the multiplicative identity of the ring $F[x]$, that is:
$f(x)\cdot 1 = 1\cdot f(x) = f(x)$, for all $f(x) \in F[x]$.
If $a \neq 0 \in F$, by the very definition of a field, $a$ is a unit in $F$: we have $U(F) = F -\{0\}$.
Note that the map $i: F \to F[x]$ given by $i(a) = a$ is an injective ring-homomorphism. This means, essentially, that "constant polynomials" are a COPY of the field $F$ inside the polynomial ring $F[x]$.
Since $i(F)$ is thus *isomorphic* to $F$, it maps units to units. This isomorphism $i$ is "so simple" it's practically transparent, like the emperor's new clothes. In other words, if I instruct you to add $2$ to $x + 4$, and you say, "wait, the NUMBER $2$, or the (constant) polynomial $2$?", I might just smack you.
A final example: in $\Bbb Q[x]$, we have $g(x) = 4$ is a unit. Why? Because the constant polynomial $f(x) = \dfrac{1}{4}$ is also in $\Bbb Q[x]$ and:
$f(x)g(x) = \dfrac{1}{4}\cdot 4 = 1$ (although our product is the "constant polynomial' $1$, remember we IDENTIFY $1$ and $i(1)$ because $i$ is an isomorphism).
Here is another way to look at it:
consider the evaluation map $\phi_0: F[x] \to F$, This is a ring homomorphism, so:
$\phi_0(f(x)g(x)) = \phi_0(f(x))\phi_0(g(x))$, that is:
$(fg)(0) = f(0)g(0)$.
If $fg = 1$, then $f(0)g(0) = 1$, so the constant terms of $f$ and $g$ are units in $F$ (non-zero). We multiply the constant terms of $f$ and $g$ together to get the constant term of $fg$.
Note that $\phi_0 \circ i = \text{id}_F$, the identity map on $F$, and that $i \circ \phi_0$ is the identity map on $i(F)$.
Now, if $f(x)$ "has $x$'s in it", it is NOT invertible (not a unit of $F[x]$). Mostly this is because $\dfrac{1}{x} \not\in F[x]$). But the "rigorous" way to argue is:
If $\text{deg}(f) > 0$ and $fg = 1$, then $\text{deg}(fg) > 0$. But $\text{deg}(1) = 0$, contradiction.
Since $0$ (the $0$-polynomial) is never a unit (in ANY ring), the only possible units in $F[x]$ are the 0-degree polynomials, that is, the elements of $F^{\ast}$. In short:
$U(F[x]) = U(F)$.
