Is $M=\{p\in \mathcal P^3: p(x)=p(x-1), \forall x \in \Bbb R\}$ subspace of $\mathcal P^3$ and if it is, find it's basis.

You have $M=\mathbb R$, so any number will be a basis. For instance $\{1\}$ is a basis. Or $\{7\sqrt2\}$, or any other number.

If $p$ is a degree three polynomial with $p(x)=p(x-1)$, we have \begin{align} p(x)=ax^3+bx^2+cx+d &= a(x-1)^3+b(x-1)^2+c(x-1)+d \\ \ \\ &= a(x^3-3x^2+3x-1)+b(x^2-2x+1)+c(x-1)+d\\ \ \\ &=ax^3 +(-3a+b)x^2+(3a-2b+c)x-a+b-c+d. \end{align} From $-3a+b=b$, we get $a=0$. From $3a-2b+c=c$, we get $b=0$. And from $-c+d=d$, we get $c=0$. So $p(x)=d$ is constant.

Here's an even easier way. From $p(x)=p(x-1)$, we get that $p'(x)=p'(x-1)$, $p''(x)=p''(x-1)$, etc. for all derivatives. Comparing the second derivatives, we get $$ 6ax=6a(x-1)$$ for all $x$, which implies $a=0$. Now comparing second derivatives, $2bx=2b(x-1),$ implying that $b=0$. In a similar fashion it follows that $c=0$.

A polynomial is equal to $0$ if and only if all of its coefficients are equal to $0$. From your equation $$3ax^2-x(3a-2b)+3a+b-c=0$$ we obtain:

$$3a = 0$$ $$3a - 2b = 0$$ $$3a + b - c = 0$$

Hence, $a = b = c = 0$. We have no condition on $d$, so $d \in \mathbb{R}$.

So, $M$ is the space of constant polynomials. Its basis is for example $\{1\}$.