The answer to your first question is yes, but the defining equations seem harder to get (as @abx points out).
Let me explain why the answer to your question is yes.
Let $p_1 = [1,0,0,0,0], p_2 = [0,1,0,0,0], \ldots , p_5=[0,0,0,0,1]$ be five points in $\mathbb{P}^4$. Consider the line $L_{i,j}$ joining $p_i$ to $p_j$. The curve:
$$C_0 = L_{1,2} \cup L_{2,3} \cup L_{3,4} \cup L_{4,5} \cup L_{5,1}$$ is the classical pentagonal configuration (it was studied in details by Ellinsgrud-Laksov for instance).
For $a \in \mathbb{C}$, consider the following quadrics in $\mathbb{P}^4$:
$$Q_0 : ax_0^2 + a^2x_2x_3 -x_1x_4 = 0,$$
$$ Q_1 : ax_1^2 + a^2x_3x_4 - x_2x_0 =0,$$
$$ Q_2 : ax_2^2 + a^2x_4x_0 -x_3x_1 = 0,$$
$$Q_3 : ax_3^2 + a^2x_0x_1 - x_4x_2 = 0, $$
$$Q_4 : ax_4^2 + a^2x_1x_2 - x_0x_3 = 0.$$
For $a=0$, the intersection of this five quadrics is the penthagonal configuration $C_0$ and for $a$ generic, you can prove (with some computations) that intersection of these quadrics is scheme-theoretically a normal elliptic quintic in $\mathbb{P}^4$.
Hence you get a smoothing of $C_0$ in $\mathbb{P}^4$ by taking $X$ the intersection of the five quadrics $\{Q_i\}_{i=0\ldots 4}$ in $\mathbb{P}^4 \times \mathbb{A}^1$ and $\pi : X \longrightarrow \mathbb{A}^1$ the natural projection.
Now, your question is related to a smoothing of the penthagonal configurations in $\mathbb{P}^3$. Geometrically the smoothing is esay to obtain. Let $P$ be a generic point in $\mathbb{P}^4$. Let:
$$p_P : \mathbb{P}^4 \longrightarrow \mathbb{P}^3$$
be the projection from $P$. If $P$ is outside the secant-threefold of $C_0$, then $p_P$ realizes an isomorphism from $C_0$ to $p_P(C_0)$, the latter being a pentagonal configuration in $\mathbb{P}^3$.
The projection map $p_P : C_a \longrightarrow p_P(C_a)$ is then an isomorphism for generic $a \in \mathbb{A}^1$ and the curve $p_P(C_a)$ is a quintic elliptic curve in $\mathbb{P}^3$.
One obtains the smoothing you are longing for in the following way: let $\mathcal{C}$ be the projection of $X$ in $\mathbb{P}^3 \times \mathbb{A}^1$. The projection map $\pi : \mathcal{C} \longrightarrow \mathbb{A}^1$ staisfies $\pi^{-1}(0) = P_p(C_0)$ is a pentagonal configuration and for $a \in \mathbb{A}^1$, the curve $\pi^{-1}(a)$ is a smooth elliptic quintic curve in $\mathbb{P}^3$.
The second part of your question is more difficult (at least in $\mathbb{P}^3$). It is easy in $\mathbb{P}^4$ because it is easily proved that $\mathcal{J}_C(2)$ is globally generated, where $C$ is a normal elliptic quintinc. On the other hand, it is easily shown that a quintic elliptic curve in $\mathbb{P}^3$ is contained in at most one quadric surface (exactly one when it is the nodal/cuspidal projection of a normal elliptic curve in $\mathbb{P}^4$, and probably none when it is generic). Hence the equations of the smoothing can't be quadratic, whch makes the situation more difficult. I don't know if cubic equations are sufficient...
EDIT : Since I found it a fun "homework" I kept on working on it, and I think one can explicit the equations with Macaulay2. In fact, getting the equations of an elliptic curve in $\mathbb{P}^3$ is not the problem. A smooth elliptic quintic in $\mathbb{P}^3$ is scheme-theoretically defined by 5 cubics and these cubics can be found in a few seconds by Macaulay2 (using elimination theory). The hard part is to get the equations for the deformation to the pentagonal configuation. So I changed a bit my inital pentagonal configuration in $\mathbb{P}^4$.
Let $q_1 = [1,0,0,0,0], q_2 = [0,1,0,0,0], q_3=[0,0,1,0,0] , q_4=[0,0,0,1,0], q_5=[1,1,1,1,-1]$. Consider the line $D_{i,j}$ joining $q_i$ to $q_j$. The curve:
$$C_0 = D_{1,2} \cup D_{2,3} \cup D_{3,4} \cup D_{4,5} \cup D_{5,1}$$ is again a pentagonal configuration.
For $[a,b] \in \mathbb{P}^1$, consider the following quadrics in $\mathbb{P}^4$:
$$T_0 : ab(x_0+x_4)^2 + a^2(x_2+x_4)(x_3+x_4) -2b^2(x_1+x_4)x_4 = 0,$$
$$ T_1 : ab(x_1+x_4)^2 + 2a^2(x_3+x_4)x_4 - b^2(x_2+x_4)(x_0+x_4) =0,$$
$$ T_2 : ab(x_2+x_4)^2 + 2a^2x_4(x_0+x_4) -b^2(x_3+x_4)(x_1+x_4) = 0,$$
$$T_3 : ab(x_3+x_4)^2 + a^2(x_0+x_4)(x_1+x_4) -2b^2x_4(x_2+x_4) = 0, $$
$$T_4 : 4abx_4^2 + a^2(x_1+x_4)(x_2+x_4) -b^2(x_0+x_4)(x_3+x_4) = 0.$$
Let $X$ be the scheme defined by the intersection of the $T_i$ in $\mathbb{P}^4 \times \mathbb{P}^1$. The projection map $\pi : X \longrightarrow \mathbb{P}^1$ gives you (generically) a smoothing of $C_0$. Again, we can project everything into $\mathbb{P}^3$ from a generic point to get a smoothing to an elliptic curve of the pentagonal configuration you started with.
As a matter of fact, the point $[0,0,0,0,1]$ is generic enough to get that smoothing. If $p : \mathbb{P}^4 \times \mathbb{P}^1 \longrightarrow \mathbb{P}^3 \times \mathbb{P}^1$ is the projection from $[0,0,0,0,1]$, the equations of $p(X)$ are given by the ideal obtained from $I = (Q_0,Q_1,Q_2,Q_3,Q_4)$ and eliminating $x_4$. This can be done using Macaulay2. There are only $42$ equations($42$, oviously!), but they are a bit long. I can't print them all on mathoverflow, but at least, I can give you the first one (Macaualay2 does not siplify much the coefficients)
$$2a^3bx_0^3+ab^3x_0^3-2a^4x_0^2x_1-3a^2b^2x_0^2x_1-ab^3x_0^2x_1+a^3bx_0x_1^2+3a^2b^2x_0x_1^2-a^3bx_1^3-2ab^3x_1^3+3a^2b^2x_0^2x_2-ab^3x_0^2x_2-2a^3bx_0x_1x_2+2b^4x_0x_1x_2+a^3bx_1^2x_2+2ab^3x_1^2x_2+a^3bx_0x_2^2-3a^2b^2x_0x_2^2-2b^4x_0x_2^2+a^3bx_1x_2^2+2ab^3x_1x_2^2-a^3bx_2^3-2ab^3x_2^3-2a^3bx_0^2x_3-ab^3x_0^2x_3+2a^4x_0x_1x_3+2ab^3x_0x_1x_3+a^3bx_1^2x_3-3a^2b^2x_1^2x_3-2b^4x_1^2x_3+2a^4x_0x_2x_3+2ab^3x_0x_2x_3-2a^3bx_1x_2x_3+2b^4x_1x_2x_3+a^3bx_2^2x_3+3a^2b^2x_2^2x_3-2a^3bx_0x_3^2-ab^3x_0x_3^2+3a^2b^2x_1x_3^2-ab^3x_1x_3^2-2a^4x_2x_3^2-3a^2b^2x_2x_3^2-ab^3x_2x_3^2+2a^3bx_3^3+ab^3x_3^3$$
If you are really interested by the specific equations, you can contact me in PM and I will send them to you.
