No, this is not true in general.

Let $g = x+\alpha$, let $t= \beta$ be an invertible element . Then we need to find $t'$ in $R[x]$ such that $(x+\alpha) t' = \beta (x+\alpha)^j = (x+ \beta \alpha \beta^{-1})^j \beta $

By formal division, this implies some identity relating $\alpha$ and $\beta \alpha \beta^{-1}$.

If $R$ is the group algebra of a free group on two generators $\alpha$ and $\beta$, say, then this formal identity won't be satisfied for any $j$. To check this, just observe that the subgroup generated by $\alpha$ and $\alpha' = \beta \alpha \beta^{-1}$ is also free, and hence maps to the free abelian group on those two generators, but for $R$ a commutative ring$, $(x+\alpha)$ does not divide $(x+ \alpha')^j$ unless $\alpha'-\alpha$ is nilpotent.

No, this is not true in general.

Let $g = x+\alpha$, let $t= \beta$ be an invertible element . Then we need to find $t'$ in $R[x]$ such that $(x+\alpha) t' = \beta (x+\alpha)^j = (x+ \beta \alpha \beta^{-1})^j \beta $, so $(x+\alpha) (t' \beta^{-1}) =(x+\beta \alpha \beta^{-1})^j$.

In particular, take $R$ to be a matrix algebra with $\alpha = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $\beta = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, so $\beta \alpha \beta^{-1}$ is $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$.

If $t' \beta^{-1}$ satisfies this identity, then it must be a diagonal matrix, as any non-diagonal matrix polynomial will have nonzero off-diagonal entries when multiplied by $x+\alpha$ (by looking only at the highest-degree non-diagonal entries and multiplying them by $x$ to get the highest-degree non-diagonal entries of the product, we can see that there is no cancellation in the top degree).

So $t'$ if it exists is a diagonal matrix, which implies that $x+0$ divides $(x+1)^j$ in the ring of polynomials in one variable, which is false, and thus $t'$ does not exist.