I am going to assume that by "algebra" you simply mean a ring.

The answer is "no", in general. For example $\mathbb{Z}/5\mathbb{Z}$ is not the unit group of a ring. Indeed, suppose it was the unit group of a ring $R$, let $u\in R$ denote a generator of the unit group. Now $-1$ is always a unit, and has order dividing $2$. But it is supposed to live in a cyclic group of order $5$, which forces its order to be $1$, i.e. $1=-1$ in $R$, so that $R$ has characteristic $2$. But then $u$ generates a subring of $R$ that is a quotient of $\mathbb{F}_2[\mathbb{Z}/5\mathbb{Z}]\cong \mathbb{F}_2[x]/(x^5-1)\cong \mathbb{F}_2\times \mathbb{F}_{16}$. Since that quotient has at least $5$ distinct elements, it must be at least all of $\mathbb{F}_{16}$, so that $R^\times$ contains $\mathbb{F}_{16}^\times\cong \mathbb{Z}/15\mathbb{Z}$ — contradiction.

In general, this argument shows that a group of prime order $p$ is isomorphic to the unit group of a ring if and only if $p=2$ or of the form $2^n-1$ for some $n$.