Are there equations which have solutions in all groups but which are not algebraicly solvable I am not sure exactly how to phrase this problem so I appologise if it is not clear, also this is somewhat long but I wanted to explain exactly where I was with the problem. If you have any questions feel free to ask.

Description of Problem
Given a set of variables $\{x,y,z,...\}$ and a variable $o$ is it possible to define a finite product of these variables and their inverses $\sigma(x,y,z,...,x^{-1},y^{-1},z^{-1},...,o,o^{-1})$ (i.e. a finite sequence made up of these variables and their inverses) such that;
1) For any group $G$ and any assignment of values from $G$ to $\{x,y,z,...\}$ there exists a unique element $g$ of $G$ such that if $o$ is set to $g$;
$\sigma(x,y,z,...,x^{-1},y^{-1},z^{-1},...,o,o^{-1})=1$
and
2) There does not exist a finite product $\gamma(x,y,z,...,x^{-1},y^{-1},z^{-1},...)$ such that;
$o=\gamma(x,y,z,...,x^{-1},y^{-1},z^{-1},...)$ for all groups $G$

Rough explaination as to why I am asking here
My intuition is no but I am unsure how to prove this. There are clearly examples of equations like these solvable in all groups (i.e. $xo=1$) but these have algebraic solutions (in that example $o=x^{-1}$) and there are examples of these equations which are solvable in wide classes of groups (i.e. $o^{n!+1}x=1$ is solvable in any group of order less than $n$ with $o=x^{-1}$) but these are not solvable in all groups. In addition some equations are solvable in all groups but not uniquely (i.e. $o^2=1$ has many solutions in groups with elements of order 2 but can always be solved with $o=1$)

Progress on proof (or proof of falsehood)
It can be shown that $\sigma$ must contain exactly $\pm1$ total occurences of $o$ (where $o^{-1}$ counts as $-1$ occurence of o) using the following argument.
If $G$ is abelian then $\sigma$ can be written as $Ao^n$ for some $A$ which is a product of the other variables. For this to be solvable $o^n=A^{-1}$ must be solvable in every abelian group. If $A=1$ $o$ is not uniquely defined for groups of order $|n|$. If $A\neq1$ and $|n|\neq1$ then $o$ is not defined for groups of order $|n|$ or $n=0$ and so $o$ is not unique. Therefore $|n|=1$ and so the total number of occurences of $o$ in $\sigma$ must be $\pm1$.
In addition it is clear that there must be an odd number of occurences of $o$ greater than $1$ (this time counting $o^{-1}$ as $1$ occurence). This follows as otherwise there is a clear definition of $\gamma$ (if there is $1$ occurence) or the observation above is violated (if there are an even number of occurences).
This is where I am and I am not sure how to proceed. Appologies again for this being overly long. Any information or advice would be appreciated.
 A: If you require uniqueness of your solution, then I don’t believe this is possible.
To shorten notation, for a set $X$ I’ll write $\sigma(X)$ to denote a word in elements of $X$. Let $X$ be a set, let $o$ be a variable, and let $\sigma(X, o)$ be any word. Then the group
$$G = \langle X, s, t~|~\sigma(X, s) = \sigma(X, t) = 1\rangle$$
fails the uniqueness condition for solutions to $\sigma(X, o)$. 

So, let’s consider the case where we don’t assume uniqueness. In this case, it is (uninterestingly) possible.
The requirement that $\sigma(X, o) = 1$ for all groups means that, in particular, this must be true for the free group $F(X\cup\{o\})$. This means that $\sigma(X,o)$ must reduce to a trivial word by the definition of free groups.
The only next requirement is that $o\notin \langle X\rangle$. Thus, we can produce a situation you want in the following way: let $G$ be any group with identity element $e$, and $\sigma(X,o)$ be any word that reduces to the trivial word.
Then, the assignment $x\mapsto e$ for all $x\in X$ and $o\mapsto g$ for any $g\ne e$ in $G$ satisfies your requirements.
A: Okay so I have no answered my question. If this doesn't seem like an answer it is probably due to me being unable to clearly describe the question I wanted to ask so sorry. Also I'm not sure how best to explain the answer.

It can be shown that no such equation exists by contradiction. First assume there exists $\sigma$ as is required by the problem. Let $n$ be the number of variables it takes (excluding $o$ and only counting $x$ not $x^{-1}$ for example). Take the group $F_n$ (the free group with $n$ generators) and assign to each of the variables that $\sigma$ takes $x_1,x_2,...$ a different one of the generators $g_1,g_2,...$. Find the value of $o$ corresponding to this situation. By definition of $F_n$, $o$ can be written as a finite product of $g_1,g_2,...$. Write $\gamma$ as this representaion of $o$ but where the generators are replaced by the corresponding variables. Therefore by definition of the free group in all groups $o=\gamma$ is a solution.

Sorry if this wasn't clear. Also thanks to Santana Afton for mentioning using the free group in there answer.
