The topology induced by a family $\mathscr{P}$ of seminorms on the vector space $X$ always makes $X$ a topological vector space (unless, as is frequently done, one requires a topological vector space to be Hausdorff, then one needs a separating family of seminorms, otherwise the topology will not be Hausdorff; but addition and scalar multiplication are always continuous with respect to the topology induced by $\mathscr{P}$).
For addition, the triangle inequality alone ensures that. Fix a point $(x_0,y_0) \in X\times X$, then for arbitrary $x,y$ we have
$$p\bigl((x+y) - (x_0+y_0)\bigr) = p\bigl((x-x_0) + (y-y_0)\bigr) \leqslant p(x-x_0) + p(y-y_0)\tag{1}$$
for every seminorm $p$. In the topology induced by $\mathscr{P}$, a neighbourhood basis of $\xi\in X$ is given by the sets
$$V(\xi;p_1,\dotsc,p_n;\varepsilon) = \bigcap_{k = 1}^n \{ \eta \in X : p_k(\eta - \xi) < \varepsilon\},$$
where $n\in\mathbb{N}$, $p_k\in\mathscr{P}$ for $1\leqslant k \leqslant n$, and $\varepsilon > 0$.
Given a neighbourhood $V(x_0+y_0;p_1,\dotsc,p_n;\varepsilon)$ of $x_0 + y_0$, from $(1)$ it follows that
$$U\times W \subset S^{-1}(V(x_0+y_0;p_1,\dotsc,p_n;\varepsilon))$$
with the neighbourhoods $U = V(x_0;p_1,\dotsc,p_n;\varepsilon/2)$ of $x_0$ and $W = V(y_0;p_1,\dotsc,p_n;\varepsilon/2)$ of $y_0$. Hence the addition $S\colon X\times X \to X$ is continuous at $(x_0,y_0)$.
Since $V(\xi;p_1,\dotsc;p_n;\varepsilon) = \xi + V(0;p_1,\dotsc,p_n;\varepsilon)$, one sees that the addition is even uniformly continuous.
For the continuity of the scalar multiplication $M \colon \mathbb{C}\times X \to X$, we also use the homogeneity of seminorms. Fixing $(\lambda_0,x_0) \in \mathbb{C}\times X$, we write
$$\lambda x - \lambda_0 x_0 = (\lambda - \lambda_0)x_0 + \lambda_0(x - x_0) + (\lambda - \lambda_0)(x - x_0),$$
whence
$$p(\lambda x - \lambda_0 x_0) \leqslant \lvert \lambda - \lambda_0\rvert p(x_0) + \lvert \lambda_0\rvert p(x-x_0) + \lvert\lambda - \lambda_0\rvert p(x-x_0)$$
for every seminorm $p$. Then $\lvert\lambda - \lambda_0\rvert < \dfrac{\min\:\{\varepsilon, 1\}}{3(1+p(x_0))}$ and $p(x-x_0) < \dfrac{\min\:\{\varepsilon, 1\}}{3(1+\lvert\lambda_0\rvert)}$ imply
$$p(\lambda x - \lambda_0 x_0) < \frac{\varepsilon}{3}\frac{p(x_0)}{1+p(x_0)} + \frac{\varepsilon}{3}\frac{\lvert\lambda_0\rvert}{1+\lvert\lambda_0\rvert} + \frac{\varepsilon}{3} \frac{\lvert\lambda_0\rvert p(x_0)}{3(1+\lvert\lambda_0\rvert)(1+p(x_0))} < \varepsilon.$$
Taking intersections over finite subfamilies of $\mathscr{P}$, we see that scalar multiplication is continuous at $(\lambda_0,x_0)$.
It is worth noting that, unless $\mathscr{P}$ contains only the trivial seminorm, the scalar multiplication is not uniformly continuous.
Best Answer
The definition of (sequentially) complete is "every Cauchy sequence has a limit", not "every Cauchy sequence has a unique limit". So the fact that the space is non-separated is irrelevant to its completeness.