Are there equations which have solutions in all groups but which are not algebraicly solvable

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 $\sigma(X, o)$.

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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.