Let $S$ be a finite set and $\mathbb{K}$ be a field.
Now, if $\mathcal{F}(S)$ is the free $\mathbb{K}$-algebra generated by $S$, and $\mathcal{R}_S$ is the smallest two-sided ideal containing a set of relations $R_S$ (each relation is a non-commutative polynomial of elements $s\in S$), then we have a finitely a presented $\mathbb{K}$-algebra $\mathcal{F}(S)/\mathcal{R_S}$.
Suppose that $f$ is a one-to-one non-commutative $\mathbb{K}$-polynomial in the elements of $S$ and consider the finitely presented $\mathbb{K}$-algebra $\mathcal{F}(f(S))/\mathcal{R}_{f(S)}$, where the relations are the same non-commutative polynomials but now with the elements $f(s)\in T$.
Question: In what sense, if any, are $\mathcal{F}(f(S))/\mathcal{R}_{f(S)}$ and $\mathcal{F}(S)/\mathcal{R}_S$ related as $\mathbb{K}$-algebra?
Once you realize that any map $\phi$ on $\mathcal{F}(S)$, which specifies the image of each $s\in S$ (i.e. the generators), gives a $\mathbb{K}$-algebra homomorphism if and only if $\phi(r)=0$ for all $r\in R_S$. From here it's clear the algebras are isomorphic via $f$ as mentioned in the comments.