Here are examples in all characteristics of fields with the same prime field but without a coproduct.
Let $k=\mathbb F_p$ or $\mathbb Q$ and let $ k(x)$ and $ k(y)$ be two copies of the field of rational functions over $k$.
I claim they have no coproduct in Fld.
Indeed, let $C$ be a hypothetical coproduct $C=k(x) \sqcup k(y)$.
Consider the identity morphisms $i_1:k(x)\to k(t) $ and $i_2:k(x)\to k(t)$ (where $k(t)$ is yet another copy of the rational function field over $k$)
It gives rise to a morphism $i_1\sqcup i_2:C\to k(t)$.
Since this morphism is injective, we necessarily have $trdeg_k C=1$.
However by considering the obvious morphisms $j_1:k(x)\to k(x,y) $ and $j_2:k(y)\to k(x,y) $ we obtain a field morphism $j_1\sqcup j_2:C\to k(x,y)$ whose image must contain $x$ and $y$.
In other words the (automatically injective) morphism $j_1\sqcup j_2:C\to k(x,y)$ is surjective and thus is an isomorphism, which contradicts the preceding assertion that $trdeg_k C=1$.
This implies that $k(x)\otimes_k k(y)$is not a field, which of course can also proved directly. (The slickest way being to use a theorem of Grothendieck implying that $k(x)\otimes_k k(y)$ has Krull dimension one: see here)
Edit Let me show, as an answer to a comment below, that $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 2)$ have no coproduct.
Let $C=\mathbb Q(\sqrt 2)\sqcup \mathbb Q(\sqrt 2)$ be their hypothetical coproduct in Fld.
The identities $i_1,i_2:\mathbb Q(\sqrt 2) \to \mathbb Q(\sqrt 2)$ lead to a field morphism $i_1 \sqcup i_2:C\to \mathbb Q(\sqrt 2)$.
This proves that $C=\mathbb Q(\sqrt 2)$.
Let now $j_1,j_2:\mathbb Q(\sqrt 2)\to Q(\sqrt 2)\sqcup \mathbb Q(\sqrt 2)=C=\mathbb Q(\sqrt 2)$ be the structural morphisms for the coproduct ($j_1,j_2\in Gal(\mathbb Q(\sqrt 2)/\mathbb Q))$.
Consider then the morphisms $u_1=j_1:\mathbb Q(\sqrt 2)\to \mathbb Q(\sqrt 2)$ and $u_2:\mathbb Q(\sqrt 2)\to \mathbb Q(\sqrt 2)$ where $u_2$ is fiendishly chosen as the automorphism of $Gal(\mathbb Q(\sqrt 2)/\mathbb Q))$ different from $j_2$.
It is now impossible to choose $u:C=\mathbb Q(\sqrt 2)\to \mathbb Q(\sqrt 2)$ with $u\circ j_1=u_1$ and $u\circ j_2=u_2$.
Hence the coproduct $C$ actually cannot exist.
The coproduct of rings is somewhat similar to the coproduct of monoids or groups (aka free products), but that you use $\otimes_\mathbb{Z}$ instead of $\times$ (in fact, both are special cases of a general construction which works for monoid objects in cocomplete monoidal categories).
The coproduct of two rings $R,S$ is constructed as follows: Let $|R|$ and $|S|$ denote the underlying $\mathbb{Z}$-modules. Then consider the direct sum of tensor products
$\mathbb{Z} \oplus |R| \oplus |S| \oplus \bigl(|R| \otimes_\mathbb{Z} |S|\bigr) \oplus \bigl(|S| \otimes_\mathbb{Z} |R|\bigr) \oplus (|R| \otimes_\mathbb{Z} |S| \otimes_\mathbb{Z} |R|) \oplus \bigl(|S| \otimes_\mathbb{Z} |R| \otimes_\mathbb{Z} |S|\bigr) \oplus \dotsc$
Now we mod out the relations $x_1 \otimes \dotsc \otimes x_n \equiv x_1 \otimes \dotsc \otimes x_n \otimes 1 \equiv 1 \otimes x_1 \otimes \dotsc \otimes x_n$ and $ \dotsc \otimes x_i \otimes 1 \otimes x_{i+1} \otimes \dotsc \equiv \dotsc \otimes x_i x_{i+1} \otimes \dotsc $. The quotient has a multiplication induced by
$$(x_1 \otimes \dotsc \otimes x_n ) \cdot (y_1 \otimes \dotsc \otimes y_m) := $$
$$\left\{\begin{array}{ll} x_1 \otimes \dotsc \otimes x_n \otimes y_1 \otimes \dotsc \otimes y_m & x_n \in R, y_1 \in S \text{ or } x_n \in S, y_1 \in R \\ x_1 \otimes \dotsc \otimes x_n y_1 \otimes \dotsc \otimes y_m & x_n,y_1 \in R \text{ or } x_n,y_1 \in S\end{array}\right.$$
We obtain a ring $R \sqcup S$ with obvious homomorphism $R \rightarrow R \sqcup S \leftarrow S$. It is the coproduct: Given $f : R \to T$ and $g : S \to T$, we define $h : R \sqcup S \to T$ by mapping, for example $x_1 \otimes x_2 \otimes x_3 \in |R| \otimes |S| \otimes |R|$ to $f(x_1) \cdot g(x_2) \cdot f(x_3)$. This is clearly a homomorphism of abelian groups on the infinite direct sum, but it respects the relations and therefore extends to a homomorphism on the quotient. It is checked that this is a homomorphism of rings; the unique one satisfying $h|_R = f$ and $h|_S =g$.
The construction of $R \sqcup S$ is somewhat complicated, but notice that its elements are just formal sums of products taken from $R$ or $S$, for example $r_1 + s_1 \cdot r_2 - r_3 \cdot s_2 \cdot r_4$. Actually this is how one comes up with the construction.
As with all algebraic structures, coproducts can also be described using generators and relations: If $R \cong \mathbb{Z}\langle X \rangle / I$ and $S \cong \mathbb{Z} \langle Y \rangle / J$ (for sets of variables $X,Y$ and ideals $I,J$), then $R \sqcup S = \mathbb{Z} \langle X \sqcup Y \rangle / (I+J)$. This is much easier than the construction above, but less concrete, especially when we don't have canonical presentations.
Your idea, using the forgetful functor $U : \mathsf{Ring} \to \mathsf{Mon}$, can also be made to work: $R \sqcup S = \mathbb{Z}[U(R) \sqcup U(S)]/I$, where the ideal $I$ is generated by the relations $(r+r') \cdot 1 \equiv r \cdot 1 + r' \cdot 1$ and $1 \cdot (s+s') \equiv 1 \cdot s + 1 \cdot s'$. These relations exactly guarantee that the canonical monoid homomorphisms $U(R) \to U(\mathbb{Z}[U(R) \sqcup U(S)]) \leftarrow U(S)$ lift to ring homomorphisms $R \to \mathbb{Z}[U(R) \sqcup U(S)]/I \leftarrow S$.
Best Answer
It does exist. It is called the free product, and written $R*S$. A general element is a sum of products of elements of $R\cup S$. Since it is not required that elements commute, this cannot always be regrouped into a sum of products $rs$ with $r\in R$ and $s\in S$. Hence it is a lot larger than the tensor product.