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I put the following as answer as it does not fit within the limits of a comment, hoping that it could help. In fact, the free group is the limit of finite truncations. Below is the proposed method.

Let $k=\mathbb{F}_2$ be the usual galois field with two elements
integers, we consider

1. an infinite alphabet $X$ (denumerable is enough)
2. the set of noncommutative series $\mathcal{A}=k<<X>>>=k^{X^*}$ (i.e. all functions $X^*\rightarrow k$ with the convolution product)
3. the augmentation character $k<<X>>>\rightarrow k$ and its kernel $\frak{M}$ (series without constant term) and, for every finite subalphabet $\mathrm{F}\subset X$, the ideal $\frak{M}_\mathrm{F}$ of the series such that every monomial of the support contains at least a letter outside $\mathrm{F}$
4. the free group over $X$, $\Gamma=\Gamma(X)$
5. the group morphism $\mu : \Gamma\rightarrow 1+\frak{M}$ given by $$ (\forall x\in X)(\mu(x)=1+x) $$ which is known to be into (Magnus transformation)
6. the quotients $\mathcal{A}_{n,\mathrm{F}}=k<<X>>>/(\frak{M}^n+ \frak{M}_\mathrm{F})$ (which are finite) and the surjective quotient morphisms $q_{n,\mathrm{F}} : k<<X>>>\rightarrow \mathcal{A}_{n,\mathrm{F}}$
7. the groups $\Gamma_{n,\mathrm{F}}$, images of $q_{n,\mathrm{F}}\circ \mu$ which are finite

An interesting alternative is to take a two letter alphabet $X=\{a,b\}$, an embedding $j$ of an infinitely generated free group as the subgroup generated by the set of conjugates $\{a^nba^{-n}\}_{n\geq 0}$, set $\mathcal{A}_n=k<<X>>>/(\frak{M}^n)$ take the surjective quotient morphisms $q_n : k<<X>>>\rightarrow \mathcal{A}_n$ and replace $q_{n,\mathrm{F}}\circ \mu$ by $q_n\circ \mu\circ j$.