OP's question (v7) asks:
Is it possible to obtain a symmetric stress-energy-momentum (SEM) tensor directly from the canonical SEM tensor by adding a total derivative term to the Lagrangian? In other words, by shifting $\Delta{\cal L}=d_\mu X^\mu$, and choosing $X^\mu$ appropriately, can we exactly get the shift in the SEM tensor required, in order to make the canonical SEM tensor symmetric?
No, that project is doomed already for E&M with the Maxwell Lagrangian density
$$ {\cal L}_0~:=~ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\tag{1} $$
with
$$ F_{\mu\nu}~=~A_{\nu,\mu}-A_{\mu,\nu}, \qquad
\frac{\partial{\cal L}_0}{\partial A_{\mu,\nu}}~\stackrel{(1)}{=}~ F^{\mu\nu}.\tag{2}$$
The vacuum EL equations read
$$ 0~\approx~F^{\mu\nu}{}_{,\nu}~=~ d^{\mu}(A^{\nu}_{,\nu})-d_{\nu}d^{\nu}A^{\mu}\tag{3} $$
In E&M, the canonical SEM tensor is$^1$
$$\begin{align} \Theta^{\mu}{}_{\nu}~:=~&\delta^{\mu}_{\nu}{\cal L}_0+\left(-\frac{\partial{\cal L}_0}{\partial A_{\alpha,\mu}}+ d_{\beta}\frac{\partial{\cal L}_0}{\partial A_{\alpha,\mu\beta}}\right) A_{\alpha,\nu}
-\frac{\partial{\cal L}_0}{\partial A_{\alpha,\mu\beta}}A_{\alpha,\nu\beta}\cr
~\stackrel{(1)}{=}~&\delta^{\mu}_{\nu}{\cal L}_0+F^{\mu\alpha}A_{\alpha,\nu}~,\end{align}\tag{4}$$
while the symmetric SEM tensor is
$$ T^{\mu}{}_{\nu}~=~\delta^{\mu}_{\nu}{\cal L}_0+F^{\mu\alpha}F_{\nu\alpha}.\tag{5}$$
So the difference is$^2$
$$\begin{align} T^{\mu}{}_{\nu} -\Theta^{\mu}{}_{\nu}~\stackrel{(4)+(5)}{=}&~ F^{\mu\alpha}A_{\nu,\alpha}~=~ d_{\alpha}(F^{\mu\alpha}A_{\nu}) - \underbrace{F^{\mu\alpha}{}_{,\alpha}}_{~\approx~0}A_{\nu} \cr
~\stackrel{?}{\approx}~&\delta^{\mu}_{\nu}\Delta{\cal L}+\left(-\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu}}+ d_{\beta}\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu\beta}}\right) A_{\alpha,\nu}
-\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu\beta}}A_{\alpha,\nu\beta}\end{align}\tag{6} $$
for some total derivative term $\Delta{\cal L}=d_\mu X^\mu$, where $X^\mu$ depends on $A$ and $\partial A$. The question mark (?) in eq. (6) is OP's question. Note that the continuum equation is unaltered on-shell
$$ d_{\mu}T^{\mu}{}_{\nu} ~\approx~ d_{\mu}\Theta^{\mu}{}_{\nu}~\approx~0. \tag{7} $$
For dimensional reasons $X^\mu$ must be on the form$^3$
$$ X^{\mu}
~=~ a A^{\mu} A^{\nu}_{,\nu} + b A^{\nu} A^{\mu}_{,\nu} + c A^{\nu} A_{\nu}^{,\mu}\tag{8} $$
for some constants $a,b,c$. Then
$$\begin{align} \Delta{\cal L}~&~=~d_\mu X^\mu
~\stackrel{(8)+(10)}{=}~\Delta{\cal L}_1+\Delta{\cal L}_2,\tag{9} \cr
\Delta{\cal L}_1~&:=~a (A^{\mu}_{,\mu})^2
+ b A^{\nu}_{,\mu} A^{\mu}_{,\nu}
+ c A^{\nu}_{,\mu} A_{\nu}^{,\mu},\tag{10} \cr
\Delta{\cal L}_2~&:=~
(a+b) A^{\mu} A^{\nu}_{,\nu\mu}
+ c A^{\mu}A_{\mu,\nu}^{,\nu}~\stackrel{(3)}{\approx}~(a+b+c) A^{\mu} A^{\nu}_{,\nu\mu}.\tag{11} \end{align} $$
Consider the last term on the right-hand side of eq. (6):
$$\begin{align}\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu\beta}}A_{\alpha,\nu\beta}
&~=~\frac{\partial\Delta{\cal L}_2}{\partial A_{\alpha,\mu\beta}}A_{\alpha,\nu\beta}\cr
&~=~\frac{a+b}{2} \left(A^{\alpha} A^{\mu}_{,\alpha\nu}+A^{\mu} A^{\alpha}_{,\alpha\nu}\right) +c A^{\alpha} A^{,\mu}_{\alpha,\nu} \tag{12}\end{align}$$
Apart from the diagonal term $\delta^{\mu}_{\nu}\Delta{\cal L}_2$, the terms in eq. (12) are the only appearances of 2nd-derivatives on the right-hand side of eq. (6). We conclude that
$$ \Delta{\cal L}_2~=~0\qquad\Leftrightarrow\qquad a+b~=~0\quad\wedge\quad c~=~0.\tag{13}$$
Similar arguments shows that eq. (6) is not possible$^4$. $\Box$
--
$^1$ In eq. (4) we have indicated the canonical SEM tensor for a Lagrangian density with up to 2nd-order derivatives. Some references, e.g. Weinberg QFT, have the opposite notational conventions for $T\leftrightarrow\Theta$. Here we are using the $(-,+,+,\ldots,+)$ Minkowski sign convention.
$^2$ In formula (6) we have neglected terms in $\Delta{\cal L}$ that depends on $\partial^3A$, $\partial^4A$, $\partial^5A$, $\ldots$, etc. Such terms are excluded for various reasons.
$^3$ In retrospect, this answer completely shares the premise/ideology/program/conclusion of this Phys.SE post.
$^4$ Interestingly, if we just take the trace of eq. (6), we get
$$\begin{align} A^{\nu}_{,\mu} A^{\mu}_{,\nu}
- A^{\nu}_{,\mu} A_{\nu}^{,\mu} &~=~F^{\mu\alpha}A_{\mu,\alpha}\cr
&~\stackrel{?}{\approx}~n \Delta{\cal L}+\left(-\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu}}+ d_{\beta}\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu\beta}}\right) A_{\alpha,\mu}
-\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu\beta}}A_{\alpha,\mu\beta}\cr
&~\stackrel{(9)}{=}~(n-2) \Delta{\cal L}_1+(n-1) \Delta{\cal L}_2+ A_{\alpha,\mu} d_{\beta}\frac{\partial\Delta{\cal L}_2}{\partial A_{\alpha,\mu\beta}}\cr
&~\stackrel{(11)}{=}~(n-2) \Delta{\cal L}_1+(n-1) \Delta{\cal L}_2+ \frac{a+b}{2}\left((A^{\mu}_{,\mu})^2+A^{\nu}_{,\mu} A^{\mu}_{,\nu} \right) +c A^{\nu}_{,\mu} A_{\nu}^{,\mu} ,\tag{14}\end{align} $$
which leads to the linear eq. system
$$ \begin{align}
0&~=~a+b+c, \tag{15}\cr
-1&~=~(n-1)c\qquad\qquad\qquad\Rightarrow\qquad c~=~-\frac{1}{n-1},\tag{16}\cr
0&~=~(n-2)a +\frac{a+b}{2}\qquad\Rightarrow\qquad a~=~-\frac{1}{2(n-1)(n-2)},\tag{17}\cr
1&~=~(n-2)b +\frac{a+b}{2}\qquad\Rightarrow\qquad b~=~\frac{2n-3}{2(n-1)(n-2)},\tag{18}\end{align} $$
which remarkably has a unique & consistent solution. So it is not enough to just take the trace of eq. (6). However together with eq. (13), we conclude that there is no solution. $\Box$
Best Answer
Answering my own question
We have:
$T^{ab} = -\frac{1}{4\pi} g^{ac}F_{cd}\partial^b A^d - g^{ab}L_{elec}$
$g^{0i} = 0$. In the first term we have, (using $F_{00} = 0$):
$T^{0i} = -\frac{1}{4\pi} g^{0c}F_{cd}\partial^i A^d - g^{0i}L_{elec}$
we need $c=0$, because $g^{00}=1$:
$T^{0i} = -\frac{1}{4\pi} g^{00}F_{0d}\partial^i A^d$
with $d=i,j,k$
$F_{0d}=(E_i + E_j + E_k)$
then, whe have:
$T^{0i} = -(E_i\partial^i A^i + E_j\partial^i A^j + E_k\partial^i A^k)$
$T^{0i}= - \frac{1}{4 \pi} \left( E_x \partial^x A^x + E_y \partial^x A^y + E_z \partial^x A^z \right )$
using $- B^j = \partial_k A^i - \partial_i A^k $:
$T^{0i}= \frac{1}{4 \pi} \left(E_i \partial_i A^i + E_j \partial_j A^i + E_j B_k - E_k B_j +E_k \partial_k A^i \right ) $
with $ \nabla \cdot \vec{E} = 0 $:
$ T^{0i} = \frac{1}{4\pi} \left( ( \vec{E} \times \vec{B} )_i + \nabla \cdot (A_i \vec{E}) \right)$
QED.