Polynomial roots in the ring extension Let $R$ be a ring with identity (not necessarily commutative) and $R[x]$ be a ring of polynomials over $R$. We say that a ring $S$ is an extension of $R$ if there is a subring $\tilde{R}$ in $S$ isomorphic to $R$.
Let $S$ be an extension of $R$, and $$\phi: R\to \tilde{R}\subset S$$
be a ring isomorphism.
We say that a polynomial $f(x) = \sum\limits_{j\geq 0}f_jx^j\in R[x]$ has a root $\alpha\in S$ if
$$
\sum\limits_{j\geq 0}\phi(f_j)\alpha^j = 0.
$$
In the case where $R$ is a commutative ring, every monic polynomial $f(x)\in R[x]$ has a root $[x]_f$ in the extension $S = R[x]/R[x]f(x)$ of $R$. In the case where $R$ is not commutative, the set $R[x]/R[x]f(x)$ is a left $R[x]$-module but not a ring, because the ideal $R[x]f(x)$ is not two-sided ideal, but only one-sided. Also in non-commutative case there are examples such that two-sided ideal containing $f(x)$, that is, the ideal $R[x]f(x)R[x]$, is equal to $R[x]$, and in this case $R[x]/R[x]f(x)R[x]$ is isomorphic to zero ring.
I want to prove that for every ring with identity $R$ and every monic polynomial $f(x)$ over $R$ there exists an extension $S$ of $R$ such that $f(x)$ has a root in $S$.
 A: In the noncommutative case, your condition for a "root" is called a "right root".  I remember that T.Y. Lam worked with this condition a bit (you might search through his papers, or look in his "First Course in Noncommutative Rings" book).
It is easy to get commutative rings where your condition fails (because even though $R[x]f(x)$ is an ideal in $R[x]$, it still intersects $R$)!  For instance, let $F$ be a non-zero ring with identity and consider the ring
$$R=F[a\ :\ a^2=0].$$
The polynomial $f(x)=1+ax\in R[x]$ cannot have any root in any extension ring of $R$, because this would force $a$ to be a unit in the extension ring, but also nilpotent.

Edited to add: I've been thinking about the new question and monic polynomials.  Without loss of generality think of $R$ as contained in $S$.  If $R$ is allowed to have a different unit than $S$, I think that the answer is positive.  When $R$ is forced to be a unital subring of $S$ (i.e. containing the same unit) the answer gets a bit harder, as I'll describe below.
In the latter case, take $S:=R\coprod_{\mathbb{Z}}\mathbb{Z}[t]/(f(t))$.  Our goal is to show that $R$ is a unital subring of $S$, and we are done.  We can do that by writing elements of $S$ in reduced form.
To explain the motivation, take $f$ to be a quadratic polynomial for a moment.  Say $f(x)=x^2+bx+c$ with $b,c\in R$.  Ignoring the "bar" notation (since $S$ is a factor ring) for convenience, we have the relation
$$t^2\mapsto -bt-c.$$
We can reduce $t^3$ in two ways, and that gives us a new relation
$$tbt\mapsto -bt-tc+ct-bc.$$
We can now reduce $t^2bt$, $tbt^2$, and $tbtbt$ in two ways (each), and we get another relation beginning $tb^2t\mapsto\cdots$.  As long as $b^n\notin \mathbb{Z}$ for each $n\geq 1$, I believe that this demonstrates that your question is positively answered (for quadratics, and this can be extended).
When $b^n\in \mathbb{Z}$ things get more complicated, but the problem may still be tractable.
When $R$ is allowed to have a different identity from $S$, there is an even easier construction.
A: I have found a very simple and demonstrative construction of the ring extension, which came from non-commutative generalization of Hamilton-Caley's Theorem.
Let $R$ be a ring and $f(x) = x^m-\sum\limits_{j=0}^{m-1}f_jx^j\in R[x]$ be a monic polynomial.
We identify ring $R$ with a subring $\tilde{R} = \{\mathrm{diag}(r,r,\ldots,r): r\in R\}\subset M_m(R)$.
Then $f(x)$ has a root of the form
$$
\alpha=\left(\begin{array}{llllll}
0& e& 0&\ldots& 0& 0 \\
0& 0& e&\ldots& 0& 0 \\
. & . &  . & . & .& . \\
0& 0&0&\ldots& 0& e \\
f_0& f_1& f_2&\ldots& f_{m-2}& f_{m-1} \\
\end{array}
\right)
$$
That is $\alpha^m-\sum\limits_{j=0}^{m-1}f_j\alpha^j = 0\in M_m(R)$.
But we note that in general:
$f(\alpha^T)\neq 0$.
See article for proof  of non-commutative generalization of Hamilton-Caley's Theorem.
