Let $A$ be a unital algebra over a field $F$ with unity $1$. If $a,b,c\in A$ such that $ab=ba=1=ac=ca$, does this imply that $b=c$?
We have $c(ab)=c$. However, algebras are not necessarily associative so we can't conclude $c=c(ab)=(ca)b=b$.
Are there examples of algebras over a field with elements having multiple inverses?
Yes. Consider $S=\{e,a,b\}$ with the following multiplication table
$$ \begin{array}{c|c|c|c} & e & a & b \\\hline e & e & a & b \\\hline a & a & e & e \\\hline b & b & e & e \\ \end{array}$$
which is nonassociative with unity and with multiple inverses.
Now let $F$ be any field and consider $F[S]$ which is a vector space over $F$ with $S$ as a basis. The multiplication on $F[S]$ is induced linearly from the multipication on $S$, i.e.
$$\bigg(\sum_{i=1}^n \lambda_i s_i\bigg)\cdot \bigg(\sum_{i=1}^n\tau_i s_i\bigg):=\sum_{i,j=1}^n\lambda_i\tau_j s_is_j$$
This makes it a unital, nonassociative algebra over $F$, but $a$ has at least two inverses: itself and $b$.